Blind 75: Binary Tree Level Order Traversal #198
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Note Other AI code review bot(s) detectedCodeRabbit has detected other AI code review bot(s) in this pull request and will avoid duplicating their findings in the review comments. This may lead to a less comprehensive review. 📝 WalkthroughWalkthroughAdds a new Binary Tree Level Order Traversal problem with implementation, tests, and docs; removes per-collection progress lines and the "Coming Soon" subsection; increments GreatFrontEnd problem count to 12 in the top-level README. Changes
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Claude finished @pertrai1's task in 1m 14s —— View job Code Review: Binary Tree Level Order TraversalReviewing this PR as a coding challenge learning exercise. Progress
1. Correctness ✅The solution correctly implements level-order traversal using DFS (Depth-First Search) rather than the more common BFS approach. Let me trace through Example 2: Edge Cases Handled:
Test Coverage: Good! The test suite covers the key edge cases. However, I'd suggest one additional test: it('should handle a tree with only left children', () => {
const root: TreeNode = {
val: 1,
left: {
val: 2,
left: { val: 3, left: null, right: null },
right: null
},
right: null
};
const result = binaryTreeLevelOrderTraversal(root);
expect(result).toEqual([[1], [2], [3]]);
});2. Complexity Analysis 📊Time Complexity: O(n) where n = number of nodes
Space Complexity: O(h + n) where h = tree height
Note: The iterative BFS approach would have O(w) space for the queue where w is max tree width, which could be more space-efficient for narrow/deep trees. 3. Learning Focus: Algorithmic Patterns 🎯Pattern Used: DFS with Level Tracking This is a clever twist on the typical level-order traversal pattern. Most solutions use BFS (breadth-first search) with a queue, but this uses DFS (depth-first search) with a level parameter. Key Insight:
Why This Approach Works:
4. Code Quality 🔍Strengths:
Suggestions:
/**
* Binary Tree Level Order Traversal using DFS
* Time: O(n) - visit each node once
* Space: O(h + n) - h for call stack, n for output
*
* @param root - root node of binary tree
* @returns array of arrays, each representing a tree level
*/
export function binaryTreeLevelOrderTraversal(root: TreeNode | null): number[][] {
5. Alternative Approaches 💡Approach 1: BFS with Queue (More Common) export function binaryTreeLevelOrderTraversalBFS(root: TreeNode | null): number[][] {
if (!root) return [];
const result: number[][] = [];
const queue: TreeNode[] = [root];
while (queue.length > 0) {
const levelSize = queue.length;
const currentLevel: number[] = [];
for (let i = 0; i < levelSize; i++) {
const node = queue.shift()!;
currentLevel.push(node.val);
if (node.left) queue.push(node.left);
if (node.right) queue.push(node.right);
}
result.push(currentLevel);
}
return result;
}Trade-offs:
Approach 2: BFS with Null Markers 6. Related Problems for Practice 🔗To build pattern recognition, try these similar problems:
Pattern Family: Tree Level Traversal / BFS Problems Learning Summary 📚What Makes This Solution Interesting: Key Takeaways:
When to Use This Pattern:
Great work on implementing this with DFS! The solution is correct, efficient, and well-tested. The only enhancement would be adding complexity analysis comments to the code itself for future reference. |
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✅ Test ResultsTests have been executed for this pull request. 📊 Coverage Summary
Check the workflow logs for detailed test output. |
🧮 Solution AnalysisPR: Blind 75: Binary Tree Level Order Traversal 📊 Code Quality ResultsESLint Analysis
Issues by file:
🔍 Complexity Analysisbinary-tree-level-order-traversal📋 Analysis Results: 🔍 Automated Complexity Analysis
File: greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.ts Time ComplexityEstimated: O(1)
Space ComplexityEstimated: O(1) Data Structures Used
Code Metrics
Recommendations
binary-tree-level-order-traversal.test📋 Analysis Results: 🔍 Automated Complexity Analysis
File: greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.test.ts Time ComplexityEstimated: O(1)
Space ComplexityEstimated: O(n²) Data Structures Used
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Analysis generated for PR files only |
🧮 Solution AnalysisPR: Blind 75: Binary Tree Level Order Traversal 📊 Code Quality ResultsESLint Analysis
🔍 Complexity Analysisbinary-tree-level-order-traversal📋 Analysis Results: 🔍 Automated Complexity Analysis
File: greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.ts Time ComplexityEstimated: O(1)
Space ComplexityEstimated: O(1) Data Structures Used
Code Metrics
Recommendations
binary-tree-level-order-traversal.test📋 Analysis Results: 🔍 Automated Complexity Analysis
File: greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.test.ts Time ComplexityEstimated: O(1)
Space ComplexityEstimated: O(n²) Data Structures Used
Code Metrics
Recommendations
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📁 Analysis Files CreatedDetailed analysis files have been automatically generated and committed to 🎯 Next Steps
Analysis generated for PR files only |
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🤖 Gemini Code Review (Model: Here's a review of the provided solution for "Binary Tree Level Order Traversal": 1. Pattern IdentificationThis solution uses a Depth-First Search (DFS) pattern. Specifically, it's a recursive DFS that tracks the current The more conventional approach for level-order traversal is Breadth-First Search (BFS) using a queue. 2. Complexity Verification
3. Key InsightThe key insight for this particular DFS-based solution is to pass and maintain the current As the DFS explores the tree:
By doing this, even though the nodes are visited in a DFS order, their values are correctly grouped into their respective level arrays, effectively reconstructing the level-order output. 4. Edge Case AnalysisThe current solution handles the following edge cases correctly:
The solution appears robust for all standard tree structures and edge cases. 5. Learning Points
6. Code QualityThe code quality is good:
Minor Improvement: export function binaryTreeLevelOrderTraversal(
root: TreeNode | null
): number[][] {
const levels: number[][] = [];
// Define the helper function locally to encapsulate it
function traverseLevel(node: TreeNode, level: number): void {
if (levels.length === level) {
levels.push([]);
}
levels[level].push(node.val);
// Use a ternary operator for conciseness if preferred, or keep as is.
// The current if-block is very clear.
if (node.left) {
traverseLevel(node.left, level + 1);
}
if (node.right) {
traverseLevel(node.right, level + 1);
}
}
if (root) {
traverseLevel(root, 0);
}
return levels;
}7. Alternative Approaches
The chosen recursive DFS approach is a perfectly valid and efficient solution, demonstrating a good understanding of how to adapt DFS for level-based output. |
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🤖 Gemini Code Review (Model: Here's a detailed review of the provided solution for "Binary Tree Level Order Traversal": 1. Pattern IdentificationThis solution uses a Depth-First Search (DFS) pattern, specifically a pre-order traversal (root, then left child, then right child). It augments the standard DFS by passing an additional 2. Complexity Verification
3. Key InsightThe key insight for solving level-order traversal using a DFS approach is to pass the current depth (or level) of the node as an argument during the recursive calls. By doing so, you can determine which sub-array in your result structure ( 4. Edge Case AnalysisThe current solution handles the following edge cases correctly:
No apparent edge cases are missed by the current solution. 5. Learning Points
6. Code QualityThe code quality is good:
Minor Suggestion (more of a stylistic choice): // Current structure (good)
export function binaryTreeLevelOrderTraversal(root: TreeNode | null): number[][] {
const levels: number[][] = [];
function traverseLevel(node: TreeNode, level: number): void {
// ... logic ...
}
if (root) {
traverseLevel(root, 0);
}
return levels;
}7. Alternative Approaches
For level-order traversal, BFS is typically the go-to, but the recursive DFS solution provided is a perfectly valid and elegant alternative. |
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| if (node.left) { | ||
| traverseLevel(node.left, level + 1); | ||
| } | ||
| if (node.right) { | ||
| traverseLevel(node.right, level + 1); |
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This recursive DFS calls itself once per tree level; for a highly skewed tree (depth near the 10,000-node constraint), JavaScript/TypeScript runtimes typically throw RangeError: Maximum call stack size exceeded. That means valid inputs can crash rather than return a result. An iterative BFS/queue (or explicit stack) avoids call-stack limits while preserving level order.
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Pull request overview
This PR adds a Binary Tree Level Order Traversal solution to the Blind 75 collection using a depth-first search (DFS) approach with recursive level tracking.
Changes:
- Adds new TypeScript solution for Binary Tree Level Order Traversal with comprehensive test coverage
- Updates GreatFrontEnd problem count from 11 to 12
- Cleans up progress tracking sections in README files
Reviewed changes
Copilot reviewed 5 out of 5 changed files in this pull request and generated 2 comments.
Show a summary per file
| File | Description |
|---|---|
greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.ts |
Implements DFS-based level order traversal with helper function |
greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.test.ts |
Adds comprehensive test cases covering edge cases and various tree structures |
greatfrontend/blind-75/trees/level-order-traversal/README.md |
Documents problem description, examples, constraints, and interface |
greatfrontend/blind-75/README.md |
Updates Trees section to include new problem and cleans up progress format |
README.md |
Updates GreatFrontEnd problem count and removes outdated progress tracking |
greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.ts
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Co-authored-by: Copilot <[email protected]>
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Actionable comments posted: 5
🤖 Fix all issues with AI agents
In
`@greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.test.ts`:
- Around line 7-41: Add unit tests to binary-tree level order tests to mirror
all README examples (including cases with negative values and any edge examples)
by creating additional it(...) cases that call binaryTreeLevelOrderTraversal
with TreeNode inputs matching the README examples and asserting expected arrays;
specifically add tests for trees shown in the README (e.g., examples with
negative node values and any multi-level/unbalanced examples) so the test file
binary-tree-level-order-traversal.test.ts uses the same TreeNode shapes and
expected outputs as the README to keep documentation and implementation aligned.
In
`@greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.ts`:
- Around line 7-10: Add a comment above the binaryTreeLevelOrderTraversal
function documenting its Big O complexity: state the time complexity O(n) (where
n is number of nodes) and space complexity O(n) worst-case (for the queue and
output levels array), so place a concise comment block immediately before export
function binaryTreeLevelOrderTraversal(...) describing these complexities and
mentioning what n represents.
- Around line 12-22: The current recursive function traverseLevel causes deep
call stacks and should be replaced with an iterative BFS using a queue:
implement level-order by enqueuing the root TreeNode, then loop while queue not
empty, process nodes in each level by iterating current queue size, push
node.val into the corresponding subarray in levels, and enqueue
node.left/node.right as present; remove the recursive calls in traverseLevel (or
replace traverseLevel with an iterative levelOrder function that uses the queue)
to avoid stack overflow. Also add Big O documentation comment: "Time: O(n) |
Space: O(w)" (n = nodes, w = max width) near the function. Ensure you reference
the existing symbols TreeNode, levels, and traverseLevel (or refactor name to
levelOrder) when making the change.
In `@greatfrontend/blind-75/trees/level-order-traversal/README.md`:
- Around line 19-40: Update the README to reflect that root can be null and that
zero-node trees are allowed: change the parameter description to indicate root
may be null (e.g., "root: TreeNode | null") and add/update an example showing
Input: root = null -> Output: [] (or similar); also change the constraints from
"1 <= Number of nodes <= 10,000" to "0 <= Number of nodes <= 10,000" while
keeping the existing TreeNode.val range unchanged. Ensure references to
TreeNode.val remain accurate.
- Around line 1-5: Update the README header in level-order-traversal/README.md
to include a difficulty badge and a link to the GreatFrontEnd problem page:
insert a difficulty badge (e.g., "Difficulty: Medium") and add a hyperlink
labeled "GreatFrontEnd problem" pointing to the problem URL at the top of the
file beneath the title, ensuring the header remains concise and follows the
suggested header structure in the review comment.
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**/README.md
📄 CodeRabbit inference engine (CLAUDE.md)
Include problem title and difficulty badge, link to the problem on the respective platform (LeetCode, GreatFrontEnd, etc.), problem description, examples and constraints
Files:
README.mdgreatfrontend/blind-75/trees/level-order-traversal/README.mdgreatfrontend/blind-75/README.md
README.md
📄 CodeRabbit inference engine (AGENTS.md)
Update README.md when adding new solutions
Files:
README.md
**/*.ts
📄 CodeRabbit inference engine (.github/copilot-instructions.md)
**/*.ts: Use explicit type annotations for function parameters and return types
Follow strict TypeScript configuration (defined intsconfig.json)
Target ES2020 with CommonJS modules
Strict mode enabled with ES2020 target and CommonJS modules for TypeScript
**/*.ts: Use explicit type annotations for function parameters and return types in TypeScript solutions
Follow strict TypeScript configuration with ES2020 target and CommonJS modules
Avoidanytype unless absolutely necessary in TypeScript solutions
Use interfaces for object types in TypeScript solutions
Files:
greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.tsgreatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.test.ts
**/*.{ts,js}
📄 CodeRabbit inference engine (.github/copilot-instructions.md)
**/*.{ts,js}: Test complex algorithms with helper functions, data structure implementations, solutions with multiple edge cases, GreatFrontEnd API implementations, and reusable utility functions
Use descriptive variable names (single letters likei,j,kare acceptable for loop iterators and standard mathematical contexts)
Don't use deprecated JavaScript features
Files:
greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.tsgreatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.test.ts
**/*-*.{ts,js,md}
📄 CodeRabbit inference engine (.github/copilot-instructions.md)
Use kebab-case for file names matching problem names
Files:
greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.tsgreatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.test.ts
**/*.{js,ts}
📄 CodeRabbit inference engine (CLAUDE.md)
**/*.{js,ts}: Verify the solution correctly solves the problem for all test cases, including edge cases: empty arrays, single elements, negative numbers, null values
Document and verify time and space complexity using Big O notation, typically in code comments or docs/analysis/ files
Use clear, descriptive variable names (not just i, j, k unless in simple loops) and add comments for non-obvious logic or algorithm steps
Use const by default, let only when reassignment is needed; avoid var except for LeetCode solution function definitions
Use strict equality (===) over loose equality (==)
Use modern ES6+ syntax (arrow functions, destructuring, spread operators) where appropriate and prefer built-in methods (.map(), .filter(), .reduce())
**/*.{js,ts}: Useconstby default,letonly when reassignment is needed in JavaScript/TypeScript
Use strict equality (===) over loose equality (==) in JavaScript/TypeScript
Use descriptive variable names; single letters likei,j,kacceptable only for standard loop contexts
Prefer built-in array methods (.map(),.filter(),.reduce()) in JavaScript/TypeScript
Fix ESLint errors withnpm run quality:fixand ensure proper code formatting with Prettier
Files:
greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.tsgreatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.test.ts
**/greatfrontend/**/*.{js,ts}
📄 CodeRabbit inference engine (CLAUDE.md)
**/greatfrontend/**/*.{js,ts}: Verify correct usage of browser APIs (DOM, Fetch, Storage, etc.), check for browser compatibility considerations, and validate event handling patterns with proper cleanup
Check for unnecessary re-renders or DOM manipulations, verify efficient data structure usage for UI updates, and look for debouncing/throttling where appropriate
Verify parameter validation, error handling, and consider edge cases specific to browser environments (empty inputs, invalid types)
Files:
greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.tsgreatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.test.ts
**/*.test.{ts,js}
📄 CodeRabbit inference engine (.github/copilot-instructions.md)
Use
*.test.tsor*.test.jsnaming convention for test files
Files:
greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.test.ts
**/*.test.{js,ts}
📄 CodeRabbit inference engine (CLAUDE.md)
Use Vitest for testing complex solutions, data structures, and GreatFrontEnd implementations; ensure tests handle edge cases, async operations, and cleanup logic
Files:
greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.test.ts
**/*.test.ts
📄 CodeRabbit inference engine (AGENTS.md)
**/*.test.ts: Test all examples from problem description in test files
Add edge case tests: empty inputs, single elements, nulls, boundary conditions in test files
Use descriptive test names that explain what is being tested in test files
Group related tests usingdescribeblocks in test files
Include both positive and negative test cases in test files
Add test cases for new solutions and update tests when refactoring existing solutions
Files:
greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.test.ts
**/*-*.test.ts
📄 CodeRabbit inference engine (AGENTS.md)
Test files should use kebab-case naming with
.test.tsextension:{problem-name}.test.ts
Files:
greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.test.ts
🧠 Learnings (19)
📓 Common learnings
Learnt from: CR
Repo: pertrai1/coding-challenges PR: 0
File: AGENTS.md:0-0
Timestamp: 2025-12-30T13:41:58.622Z
Learning: Applies to leetcode/**/*.{js,ts} : Test edge cases if adding or modifying solution code
📚 Learning: 2025-12-30T13:41:24.644Z
Learnt from: CR
Repo: pertrai1/coding-challenges PR: 0
File: CLAUDE.md:0-0
Timestamp: 2025-12-30T13:41:24.644Z
Learning: Applies to **/README.md : Include problem title and difficulty badge, link to the problem on the respective platform (LeetCode, GreatFrontEnd, etc.), problem description, examples and constraints
Applied to files:
README.md
📚 Learning: 2025-12-30T13:41:07.657Z
Learnt from: CR
Repo: pertrai1/coding-challenges PR: 0
File: .github/copilot-instructions.md:0-0
Timestamp: 2025-12-30T13:41:07.657Z
Learning: Applies to **/*.{ts,js} : Test complex algorithms with helper functions, data structure implementations, solutions with multiple edge cases, GreatFrontEnd API implementations, and reusable utility functions
Applied to files:
README.mdgreatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.test.ts
📚 Learning: 2025-12-30T13:41:24.644Z
Learnt from: CR
Repo: pertrai1/coding-challenges PR: 0
File: CLAUDE.md:0-0
Timestamp: 2025-12-30T13:41:24.644Z
Learning: Applies to **/greatfrontend/**/*.{js,ts} : Verify correct usage of browser APIs (DOM, Fetch, Storage, etc.), check for browser compatibility considerations, and validate event handling patterns with proper cleanup
Applied to files:
README.md
📚 Learning: 2025-12-30T13:41:24.644Z
Learnt from: CR
Repo: pertrai1/coding-challenges PR: 0
File: CLAUDE.md:0-0
Timestamp: 2025-12-30T13:41:24.644Z
Learning: Applies to **/greatfrontend/**/*.{js,ts} : Verify parameter validation, error handling, and consider edge cases specific to browser environments (empty inputs, invalid types)
Applied to files:
README.md
📚 Learning: 2025-12-30T13:41:58.622Z
Learnt from: CR
Repo: pertrai1/coding-challenges PR: 0
File: AGENTS.md:0-0
Timestamp: 2025-12-30T13:41:58.622Z
Learning: Applies to leetcode/**/*.ts : Use TypeScript type annotations for all parameters and return types with complexity analysis in code comments
Applied to files:
greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.ts
📚 Learning: 2025-12-30T13:41:58.622Z
Learnt from: CR
Repo: pertrai1/coding-challenges PR: 0
File: AGENTS.md:0-0
Timestamp: 2025-12-30T13:41:58.622Z
Learning: Applies to leetcode/**/README.md : Keep README files up to date with problem counts and organization
Applied to files:
greatfrontend/blind-75/trees/level-order-traversal/README.mdgreatfrontend/blind-75/README.md
📚 Learning: 2025-12-30T13:41:58.622Z
Learnt from: CR
Repo: pertrai1/coding-challenges PR: 0
File: AGENTS.md:0-0
Timestamp: 2025-12-30T13:41:58.622Z
Learning: Applies to **/*.test.ts : Add edge case tests: empty inputs, single elements, nulls, boundary conditions in test files
Applied to files:
greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.test.ts
📚 Learning: 2025-12-30T13:41:58.622Z
Learnt from: CR
Repo: pertrai1/coding-challenges PR: 0
File: AGENTS.md:0-0
Timestamp: 2025-12-30T13:41:58.622Z
Learning: Applies to **/*.test.ts : Add test cases for new solutions and update tests when refactoring existing solutions
Applied to files:
greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.test.ts
📚 Learning: 2025-12-30T13:41:24.644Z
Learnt from: CR
Repo: pertrai1/coding-challenges PR: 0
File: CLAUDE.md:0-0
Timestamp: 2025-12-30T13:41:24.644Z
Learning: Applies to **/*.test.{js,ts} : Use Vitest for testing complex solutions, data structures, and GreatFrontEnd implementations; ensure tests handle edge cases, async operations, and cleanup logic
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Learning: Applies to **/*.test.ts : Test all examples from problem description in test files
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Learning: Applies to **/*.test.ts : Include both positive and negative test cases in test files
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Learning: Applies to **/*.test.ts : Group related tests using `describe` blocks in test files
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Learning: Applies to leetcode/**/*.{js,ts} : Test edge cases if adding or modifying solution code
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Learning: Applies to **/*.test.ts : Use descriptive test names that explain what is being tested in test files
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Learning: Applies to **/*.{js,ts} : Verify the solution correctly solves the problem for all test cases, including edge cases: empty arrays, single elements, negative numbers, null values
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Learning: Applies to leetcode/*/*/ : Ensure problems are in correct difficulty folders (easy, medium, hard) in the LeetCode directory
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🧬 Code graph analysis (1)
greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.test.ts (1)
greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.ts (2)
binaryTreeLevelOrderTraversal(7-30)TreeNode(1-5)
⏰ Context from checks skipped due to timeout of 90000ms. You can increase the timeout in your CodeRabbit configuration to a maximum of 15 minutes (900000ms). (2)
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- GitHub Check: review
🔇 Additional comments (2)
README.md (1)
10-10: GreatFrontEnd solved count update looks consistent.Thanks for keeping the platform totals in sync.
greatfrontend/blind-75/README.md (1)
9-40: Catalog update reads cleanly with the new Trees entry.The simplified headings and the added level-order traversal item align well.
✏️ Tip: You can disable this entire section by setting review_details to false in your review settings.
greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.test.ts
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| function traverseLevel(node: TreeNode, level: number): void { | ||
| if (levels.length === level) { | ||
| levels.push([]); | ||
| } | ||
| levels[level].push(node.val); | ||
| if (node.left) { | ||
| traverseLevel(node.left, level + 1); | ||
| } | ||
| if (node.right) { | ||
| traverseLevel(node.right, level + 1); | ||
| } |
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function recurse() {
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if (count < 15000) {
recurse();
}
}
try {
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console.log('Recursion depth reached: ' + count);
} catch (e) {
console.log('Stack overflow at depth: ' + count);
}
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Use BFS instead of recursive DFS for level-order traversal.
Recursive DFS with 10,000 nodes creates ~12,500 call frames, leaving minimal stack margin (~2.5K) before overflow. More importantly, level-order traversal is a BFS problem by definition—use an iterative queue-based approach, which is both safer and the standard pattern for this algorithm.
Additionally, add Big O complexity documentation as required by the coding guidelines.
🛠️ Proposed iterative BFS fix
export function binaryTreeLevelOrderTraversal(
root: TreeNode | null
): number[][] {
const levels: number[][] = [];
-
- function traverseLevel(node: TreeNode, level: number): void {
- if (levels.length === level) {
- levels.push([]);
- }
- levels[level].push(node.val);
- if (node.left) {
- traverseLevel(node.left, level + 1);
- }
- if (node.right) {
- traverseLevel(node.right, level + 1);
- }
- }
-
- if (root) {
- traverseLevel(root, 0);
- }
-
- return levels;
+ if (!root) {
+ return levels;
+ }
+
+ const queue: TreeNode[] = [root];
+ let index = 0;
+
+ while (index < queue.length) {
+ const levelSize = queue.length - index;
+ const currentLevel: number[] = [];
+
+ for (let i = 0; i < levelSize; i += 1) {
+ const node = queue[index];
+ index += 1;
+ currentLevel.push(node.val);
+ if (node.left) {
+ queue.push(node.left);
+ }
+ if (node.right) {
+ queue.push(node.right);
+ }
+ }
+
+ levels.push(currentLevel);
+ }
+
+ return levels;
}Time: O(n) | Space: O(w) where n is the number of nodes and w is the maximum width.
📝 Committable suggestion
‼️ IMPORTANT
Carefully review the code before committing. Ensure that it accurately replaces the highlighted code, contains no missing lines, and has no issues with indentation. Thoroughly test & benchmark the code to ensure it meets the requirements.
| function traverseLevel(node: TreeNode, level: number): void { | |
| if (levels.length === level) { | |
| levels.push([]); | |
| } | |
| levels[level].push(node.val); | |
| if (node.left) { | |
| traverseLevel(node.left, level + 1); | |
| } | |
| if (node.right) { | |
| traverseLevel(node.right, level + 1); | |
| } | |
| export function binaryTreeLevelOrderTraversal( | |
| root: TreeNode | null | |
| ): number[][] { | |
| const levels: number[][] = []; | |
| if (!root) { | |
| return levels; | |
| } | |
| const queue: TreeNode[] = [root]; | |
| let index = 0; | |
| while (index < queue.length) { | |
| const levelSize = queue.length - index; | |
| const currentLevel: number[] = []; | |
| for (let i = 0; i < levelSize; i += 1) { | |
| const node = queue[index]; | |
| index += 1; | |
| currentLevel.push(node.val); | |
| if (node.left) { | |
| queue.push(node.left); | |
| } | |
| if (node.right) { | |
| queue.push(node.right); | |
| } | |
| } | |
| levels.push(currentLevel); | |
| } | |
| return levels; | |
| } | |
| // Time: O(n) | Space: O(w) where n is the number of nodes and w is the maximum width |
🤖 Prompt for AI Agents
In
`@greatfrontend/blind-75/trees/level-order-traversal/binary-tree-level-order-traversal.ts`
around lines 12 - 22, The current recursive function traverseLevel causes deep
call stacks and should be replaced with an iterative BFS using a queue:
implement level-order by enqueuing the root TreeNode, then loop while queue not
empty, process nodes in each level by iterating current queue size, push
node.val into the corresponding subarray in levels, and enqueue
node.left/node.right as present; remove the recursive calls in traverseLevel (or
replace traverseLevel with an iterative levelOrder function that uses the queue)
to avoid stack overflow. Also add Big O documentation comment: "Time: O(n) |
Space: O(w)" (n = nodes, w = max width) near the function. Ensure you reference
the existing symbols TreeNode, levels, and traverseLevel (or refactor name to
levelOrder) when making the change.
| - `root: TreeNode`: Root node of the tree. Examples display a level-order traversal of the tree | ||
|
|
||
| ## Examples | ||
|
|
||
| ```bash | ||
| Input: root = [13,null,-55] | ||
| Output: [[13],[-55]] | ||
| Explanation: The root has two levels: the first level with the value 13, and the second level with the value -55. | ||
|
|
||
| Input: root = [1,2,3,4,5,6,7] | ||
| Output: [[1],[2,3],[4,5,6,7]] | ||
| Explanation: The tree has three levels: [1] at the first level, [2, 3] at the second level, and [4, 5, 6, 7] at the third level. | ||
|
|
||
| Input: root = [5,3,8,null,4,7,9] | ||
| Output: [[5],[3,8],[4,7,9]] | ||
| Explanation: The tree has three levels: [5] at the first level, [3, 8] at the second level, and [4, 7, 9] at the third level. | ||
| ``` | ||
|
|
||
| ## Constraints | ||
|
|
||
| - 1 <= Number of nodes <= 10,000 | ||
| - -100 <= `TreeNode.val` <= 100 |
There was a problem hiding this comment.
Align docs with null-root handling and constraints.
The implementation/tests allow null, but the input section and constraints imply at least one node. Please reflect the null case and allow zero nodes in constraints.
📝 Suggested doc fix
-- `root: TreeNode`: Root node of the tree. Examples display a level-order traversal of the tree
+- `root: TreeNode | null`: Root node of the tree. Examples display a level-order traversal of the tree
@@
-- 1 <= Number of nodes <= 10,000
+- 0 <= Number of nodes <= 10,000📝 Committable suggestion
‼️ IMPORTANT
Carefully review the code before committing. Ensure that it accurately replaces the highlighted code, contains no missing lines, and has no issues with indentation. Thoroughly test & benchmark the code to ensure it meets the requirements.
| - `root: TreeNode`: Root node of the tree. Examples display a level-order traversal of the tree | |
| ## Examples | |
| ```bash | |
| Input: root = [13,null,-55] | |
| Output: [[13],[-55]] | |
| Explanation: The root has two levels: the first level with the value 13, and the second level with the value -55. | |
| Input: root = [1,2,3,4,5,6,7] | |
| Output: [[1],[2,3],[4,5,6,7]] | |
| Explanation: The tree has three levels: [1] at the first level, [2, 3] at the second level, and [4, 5, 6, 7] at the third level. | |
| Input: root = [5,3,8,null,4,7,9] | |
| Output: [[5],[3,8],[4,7,9]] | |
| Explanation: The tree has three levels: [5] at the first level, [3, 8] at the second level, and [4, 7, 9] at the third level. | |
| ``` | |
| ## Constraints | |
| - 1 <= Number of nodes <= 10,000 | |
| - -100 <= `TreeNode.val` <= 100 | |
| - `root: TreeNode | null`: Root node of the tree. Examples display a level-order traversal of the tree | |
| ## Examples | |
🤖 Prompt for AI Agents
In `@greatfrontend/blind-75/trees/level-order-traversal/README.md` around lines 19
- 40, Update the README to reflect that root can be null and that zero-node
trees are allowed: change the parameter description to indicate root may be null
(e.g., "root: TreeNode | null") and add/update an example showing Input: root =
null -> Output: [] (or similar); also change the constraints from "1 <= Number
of nodes <= 10,000" to "0 <= Number of nodes <= 10,000" while keeping the
existing TreeNode.val range unchanged. Ensure references to TreeNode.val remain
accurate.
|
@codex review |
|
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|
Claude finished @pertrai1's task in 3m 7s —— View job Code Review: Binary Tree Level Order TraversalReviewing this PR as a coding challenge learning exercise. Progress
1. Correctness ✅The solution correctly implements level-order traversal using DFS (Depth-First Search) with level tracking rather than the more common BFS approach. Let me trace through Example 2 from the README: Edge Cases Handled:
Test Coverage Analysis: // Missing: Test for the README example with negative values
it('should handle tree with negative values', () => {
const root: TreeNode = {
val: 13,
left: null,
right: { val: -55, left: null, right: null }
};
const result = binaryTreeLevelOrderTraversal(root);
expect(result).toEqual([[13], [-55]]);
});
// Missing: Full three-level balanced tree from README
it('should handle full three-level tree', () => {
const root: TreeNode = {
val: 1,
left: {
val: 2,
left: { val: 4, left: null, right: null },
right: { val: 5, left: null, right: null }
},
right: {
val: 3,
left: { val: 6, left: null, right: null },
right: { val: 7, left: null, right: null }
}
};
const result = binaryTreeLevelOrderTraversal(root);
expect(result).toEqual([[1], [2, 3], [4, 5, 6, 7]]);
});2. Complexity Analysis 📊Documented Complexity:
Analysis: However, the space complexity documentation needs a minor clarification: More Accurate Space Complexity: O(h + n) where:
The current comment is accurate but could be clearer that the total space complexity is O(n) worst-case. Important Consideration: 3. Learning Focus: Algorithmic Patterns 🎯Pattern Used: DFS with Level Tracking This is a clever twist on the typical level-order traversal pattern. Most solutions use BFS (breadth-first search) with a queue, but this uses DFS (depth-first search) with a level parameter. Key Insight 💡The non-obvious trick is using
Why This Approach Works:
Pattern RecognitionThis pattern appears in problems requiring level-by-level grouping or processing of tree nodes. The key decision is: pass depth/level as a parameter in DFS rather than using a queue in BFS. 4. Code Quality 🔍Strengths:
Minor Observations:
No issues found - this is high-quality, production-ready code. 5. Alternative Approaches 💡Approach 1: BFS with Queue (More Common)export function binaryTreeLevelOrderTraversalBFS(root: TreeNode | null): number[][] {
if (!root) return [];
const result: number[][] = [];
const queue: TreeNode[] = [root];
while (queue.length > 0) {
const levelSize = queue.length;
const currentLevel: number[] = [];
for (let i = 0; i < levelSize; i++) {
const node = queue.shift()!;
currentLevel.push(node.val);
if (node.left) queue.push(node.left);
if (node.right) queue.push(node.right);
}
result.push(currentLevel);
}
return result;
}Trade-offs: DFS (current solution):
BFS (alternative):
Approach 2: BFS with Null MarkersYou could use 6. Related Problems for Practice 🔗To build pattern recognition with level-order traversal, try these similar problems:
Pattern Family: Tree Level Traversal / BFS-related Problems Learning Summary 📚What Makes This Solution InterestingThis problem demonstrates that level-order traversal doesn't require BFS. The DFS approach with level tracking is less intuitive but equally valid and can be more memory-efficient for certain tree shapes. Key Takeaways
When to Use This Pattern
The Core InsightThe fundamental insight is that you can achieve level-order output using depth-first traversal by maintaining a
Final Assessment ⭐Overall: This is an excellent solution - correct, efficient, well-documented, and well-tested. The choice to use DFS instead of BFS demonstrates deep understanding of tree traversal patterns. Strengths:
Minor Suggestions:
Great work on implementing this elegant DFS solution! 🎉 |
|
🤖 Gemini Code Review (Model: Here's a detailed review of your 1. Pattern IdentificationThis solution uses a Depth-First Search (DFS) pattern. Specifically, it's a recursive pre-order traversal where you process the current node, then recursively call for its left child, then its right child. The key to achieving "level order" with DFS here is the While level order traversal is most commonly implemented using Breadth-First Search (BFS) with a queue, this DFS approach is a valid and sometimes elegant alternative, especially when the output structure naturally aligns with collecting results based on depth. 2. Complexity Verification
3. Key InsightThe key insight for this DFS-based level order traversal is: "Maintain a collection of lists (or arrays) where each index corresponds to a tree level. During a standard DFS traversal, pass the current node's depth (level) as an argument. When visiting a node, ensure an array exists for its corresponding level in the collection, and then append the node's value to that level's array." The "aha!" moment is realizing that by simply tracking the 4. Edge Case AnalysisThe solution appears robust and handles the following edge cases correctly:
5. Learning Points
6. Code Quality
Minor Improvement Suggestion:
7. Alternative Approaches
Overall, this is a very well-implemented and correct solution using a less common but valid DFS approach for level-order traversal. The code quality, complexity analysis, and edge case handling are all excellent. Good job! |
|
@codex review |
|
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|
Claude finished @pertrai1's task in 3m 5s —— View job Code Review: Binary Tree Level Order TraversalReviewing this PR as a coding challenge learning exercise. Progress
1. Correctness ✅The solution correctly implements level-order traversal using DFS (Depth-First Search) with level tracking rather than the more common BFS approach. Let me trace through Example 2 from the README: Edge Cases Handled:
Test Coverage Analysis:
This is thorough and well-done! ✨ 2. Complexity Analysis 📊Documented Complexity: @time - O(n) where n is the number of nodes (visits each node once)
@space - O(h) for the recursion stack where h is the tree height, plus O(n) for the output arrayAnalysis: Time Complexity: O(n) ✅
Space Complexity: O(h + n) ✅
Note on Alternative Approaches: 3. Learning Focus: Algorithmic Patterns 🎯Pattern Used: DFS with Level TrackingThis is a clever twist on the typical level-order traversal pattern. Most solutions use BFS (breadth-first search) with a queue, but this uses DFS (depth-first search) with a level parameter. Key Insight 💡The non-obvious trick is using
Why This Approach Works:
Pattern RecognitionThis pattern appears in problems requiring level-by-level grouping or processing of tree nodes. The key decision is: pass depth/level as a parameter in DFS rather than using a queue in BFS. 4. Code Quality 🔍Strengths:
Code Style:
No issues found - this is high-quality, production-ready code! 🎉 5. Alternative Approaches 💡Approach 1: BFS with Queue (More Common)export function binaryTreeLevelOrderTraversalBFS(root: TreeNode | null): number[][] {
if (!root) return [];
const result: number[][] = [];
const queue: TreeNode[] = [root];
while (queue.length > 0) {
const levelSize = queue.length;
const currentLevel: number[] = [];
for (let i = 0; i < levelSize; i++) {
const node = queue.shift()!;
currentLevel.push(node.val);
if (node.left) queue.push(node.left);
if (node.right) queue.push(node.right);
}
result.push(currentLevel);
}
return result;
}Trade-offs: DFS (current solution):
BFS (alternative):
Approach 2: BFS with Null MarkersYou could use 6. Related Problems for Practice 🔗To build pattern recognition with level-order traversal, try these similar problems:
Pattern Family: Tree Level Traversal / BFS-related Problems Learning Summary 📚What Makes This Solution InterestingThis problem demonstrates that level-order traversal doesn't require BFS. The DFS approach with level tracking is less intuitive but equally valid and can be more memory-efficient for certain tree shapes. Key Takeaways
When to Use This Pattern
The Core InsightThe fundamental insight is that you can achieve level-order output using depth-first traversal by maintaining a
Final Assessment ⭐Overall: This is an excellent solution - correct, efficient, well-documented, and well-tested. The choice to use DFS instead of BFS demonstrates deep understanding of tree traversal patterns, and the comprehensive test coverage (including all README examples) shows attention to detail. Strengths:
What I Love:
Great work on implementing this elegant DFS solution! 🎉 |
|
🤖 Gemini Code Review (Model: Here's a review of the 1. Pattern IdentificationThis solution uses a Depth-First Search (DFS) pattern. Specifically, it employs a recursive pre-order traversal (root, left, right) combined with a 2. Complexity Verification
3. Key InsightThe key insight for solving level-order traversal using DFS is to pass the current level (depth) as a parameter during recursion and use this level to determine which sub-array in the
This way, all nodes at the same level are collected into the same inner array in the 4. Edge Case AnalysisThe solution handles the following edge cases correctly:
All common edge cases seem to be covered. 5. Learning Points
6. Code Quality
One minor point: the 7. Alternative Approaches
The current DFS recursive solution is a perfectly valid and elegant way to solve this problem, especially for those comfortable with recursion. |
Summary by CodeRabbit
New Features
Documentation
Tests
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