Binary Tree Zigzag Level Order Traversal

Data Structures
Medium
Google
111K views

Given the root of a binary tree, return the zigzag level order traversal of its nodes' values (i.e., from left to right, then right to left for the next level, and so on). Use two stacks or a deque.

Why Interviewers Ask This

Google interviewers ask this to evaluate your mastery of level-order traversal variations and your ability to manipulate data structures dynamically. They specifically test if you can adapt standard BFS logic to handle alternating directions without resorting to inefficient post-processing or excessive memory usage. It reveals your understanding of queue versus stack mechanics and how to optimize for O(N) time complexity with minimal space overhead.

How to Answer This Question

1. Clarify the input: Confirm if the tree is empty and define what constitutes a 'level'. 2. Choose your strategy: Explicitly state whether you will use two stacks (one for even, one for odd levels) or a deque with a boolean flag to reverse order. 3. Walk through the algorithm: Describe initializing the first stack with the root, then iterating while the current stack is not empty. 4. Detail the popping logic: Explain that when popping from the left stack, push children left-then-right; when popping from the right stack, push children right-then-left to achieve the zigzag effect naturally. 5. Handle edge cases: Mention handling null roots and ensuring the result list is correctly formatted. 6. Analyze complexity: Conclude by stating the O(N) time and O(N) space complexity, emphasizing why this approach is optimal for Google's performance standards.

Key Points to Cover

  • Demonstrating clear distinction between Queue-based BFS and Stack-based Zigzag logic
  • Explaining how push order determines pop order to achieve directional reversal
  • Handling the null root case gracefully before starting the loop
  • Achieving O(N) time complexity without unnecessary list reversals
  • Articulating the space trade-off clearly regarding stack depth

Sample Answer

To solve the Zigzag Level Order Traversal problem efficiently, I would start by checking if the root is null; if so, return an empty list immediately. For the core logic, I prefer using two stacks to manage the direction…

Common Mistakes to Avoid

  • Attempting to reverse the result list after a standard BFS traversal, which adds an extra pass and reduces elegance
  • Forgetting to swap the stacks or toggle the direction flag, causing all subsequent levels to traverse in the same direction
  • Pushing children in the wrong order relative to the current traversal direction, resulting in incorrect zigzag patterns
  • Neglecting to handle the empty tree case, leading to potential runtime errors during initialization

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