Level Order Traversal of a Binary Tree
Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level). Must use a Queue (BFS).
Why Interviewers Ask This
Amazon interviewers ask this to verify your grasp of Breadth-First Search and queue management, core competencies for scalable system design. They specifically test if you can handle level-by-level processing without recursion, ensuring you understand how to track tree depth dynamically while maintaining O(N) time complexity.
How to Answer This Question
1. Clarify the problem by defining inputs, outputs, and edge cases like empty trees or single nodes. 2. Propose a BFS strategy using a Queue, explicitly stating why it fits level-order requirements better than DFS. 3. Outline the algorithm: initialize the queue with the root, then loop while the queue is not empty. 4. Detail the critical step of calculating the current level size before iterating to ensure nodes are grouped correctly by depth. 5. Walk through a dry run with a small example tree, explaining how values are appended to the result list per level. 6. Analyze time and space complexity, confirming O(N) time and O(W) space where W is maximum width.
Key Points to Cover
- Explicitly mention using a Queue to enforce Breadth-First Search logic
- Demonstrate tracking current level size before iterating to separate levels correctly
- Handle edge cases like null roots or single-node trees proactively
- Correctly enqueue children (left then right) to maintain left-to-right order
- Provide clear Time and Space complexity analysis showing O(N) efficiency
Sample Answer
To solve the Level Order Traversal, I would use a Breadth-First Search approach leveraging a standard Queue data structure. First, I check if the root is null; if so, I return an empty list immediately. Otherwise, I init…
Common Mistakes to Avoid
- Using Depth-First Search instead of BFS, which fails to group nodes by level naturally
- Forgetting to calculate the queue size at the start of each loop, causing mixed levels in output
- Enqueueing children before processing the current node, leading to incorrect traversal order
- Neglecting to check for null pointers when accessing left or right children during iteration
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