I recognized the right side view part immediately and started coding before fully reading the prompt.
Use BFS level-order traversal to record the first and last node at each level, then reverse the collected lists to get bottom-to-top order. Alternatively, use DFS with depth tracking to update leftmost and rightmost nodes per level, then reverse. Clearly state time and space complexity and handle edge cases like an empty tree.
Pro tip: Mention that you can combine both left and right views in a single traversal to save time, and discuss how you'd adapt the solution if the tree is very deep (e.g., iterative BFS to avoid recursion stack overflow).
Confirm that 'visible from the right/left' means the rightmost/leftmost node at each depth, and that output should be bottom-to-top. Ask about empty tree, single node, and whether nodes can have duplicate values.
Decide between BFS (level-order) or DFS (pre-order with depth). BFS naturally gives level-by-level access; DFS can be more space-efficient for skewed trees. Explain your choice.
During traversal, record the first and last node encountered at each depth. For BFS, the first node in the level queue is leftmost, the last is rightmost. For DFS, update leftmost when visiting a depth for the first time, and rightmost whenever you visit a node at that depth (since pre-order visits rightmost last).
After traversal, reverse the lists of leftmost and rightmost nodes to satisfy the bottom-to-top requirement. Alternatively, build the lists by prepending, or traverse levels in reverse if using BFS with a stack.
State time complexity O(N) and space complexity O(N) for storing results (or O(H) for DFS recursion stack). Walk through a small example to verify correctness, including edge cases.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.