I got the BFS part pretty quickly but fumbled on the direction-flipping logic.
Use BFS with a queue to process the tree level by level, tracking the current level's size. For each level, collect node values in a list, and if the level is odd (0-indexed), reverse the list before adding to the result. This achieves O(n) time and O(w) space where w is the maximum width of the tree.
Pro tip: Mention that you can avoid reversing the list by using a deque and appending to the front or back based on the direction, which can be more efficient in practice. Also, clarify that O(w) space is optimal because the queue holds at most the width of the tree.
Confirm the zigzag order definition and discuss edge cases like empty tree, single node, and skewed tree. Ask if the output should be a list of lists.
Explain that BFS naturally processes level by level, and a queue is ideal for this. Mention that the queue size at any time is at most the tree's width, giving O(w) space.
For each level, record the number of nodes (level size), then process exactly that many nodes. Use a boolean flag to alternate direction, and either reverse the level list or use a deque to insert at front/back.
State that each node is visited once, so time is O(n). Space is O(w) for the queue, where w is the maximum width, which is optimal for BFS.
Walk through a small example (e.g., [3,9,20,null,null,15,7]) to verify. Mention alternative approaches like DFS with level tracking, but note BFS is more intuitive for level order.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.