BFS was the right call here and I knew it pretty fast.
Use BFS level-order traversal to process each level, capturing the first node for the left view and the last node for the right view. Alternatively, use DFS with depth tracking, updating the left view when visiting a level for the first time and the right view when visiting from the right side first. Clearly explain the trade-offs between BFS and DFS in terms of code simplicity and space complexity.
Pro tip: Mention that both views can be computed in a single traversal by storing the first and last nodes of each level, and discuss how to handle edge cases like skewed trees or single-node trees. This shows you think about efficiency and robustness.
Confirm the definition of left and right views, and ask about edge cases such as empty tree, single node, or skewed trees. This ensures you understand the requirements before coding.
Decide between BFS (level-order traversal) and DFS (pre-order with depth tracking). BFS is more intuitive for level-based views, while DFS can be more space-efficient for balanced trees.
For BFS, use a queue to process nodes level by level, recording the first and last node of each level. For DFS, traverse left-first for left view and right-first for right view, using a depth parameter to track the first visit at each level.
Store the left view nodes and right view nodes in separate lists as you traverse. Return both lists as the final output.
State the time complexity O(n) and space complexity O(n) for BFS (or O(h) for DFS). Walk through a small example to verify correctness and discuss potential optimizations.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.