I jumped straight to BFS which was fine, but I got a little tangled trying to return both views in one pass.
Use BFS level-order traversal to process nodes level by level, recording the first and last node at each level for left and right views respectively. 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 chosen approach, its time and space complexity, and handle edge cases like an empty tree.
Pro tip: Mention that both views can be obtained in a single traversal by tracking the first and last nodes per level, and discuss the trade-offs between BFS and DFS in terms of code simplicity and memory usage.
Confirm that the views are from top to bottom, and discuss handling of an empty tree or single-node tree. Ask if the tree is balanced or if there are constraints on node values.
Decide between BFS (level-order traversal) and DFS (pre-order with depth tracking). Explain why one might be preferred, e.g., BFS naturally processes levels in order.
For BFS: use a queue, process each level, record the first and last node values. For DFS: traverse left-first for left view and right-first for right view, using a depth parameter to know when to add a node.
State that both approaches run in O(n) time and O(n) space in the worst case (queue or recursion stack). Mention that BFS may use O(w) space where w is max width, while DFS uses O(h) where h is height.
Walk through a small example (e.g., a tree with 3 levels) to verify the algorithm produces correct left and right views. Consider edge cases like skewed trees.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.