I went with BFS and grabbed the last node at each level.
Perform a level-order traversal (BFS) using a queue, and for each level, record the last node's value. Alternatively, use DFS with depth tracking, prioritizing the right child, and record the first node encountered at each depth. Both approaches yield the right side view in O(n) time.
Pro tip: Clarify with the interviewer whether the tree can be empty or have only one node, and mention that the right side view includes nodes that are not necessarily right children but are the rightmost at their depth. This shows attention to edge cases and definition precision.
Confirm that the right side view consists of the rightmost node at each depth. Discuss edge cases: empty tree, single node, skewed tree, and nodes that are only visible from the right.
Decide between BFS (level-order) and DFS (pre-order with right-first). BFS is intuitive: process level by level and take the last node. DFS is more space-efficient for deep trees: track depth and record the first node seen at each depth.
For BFS: use a queue, iterate level by level, and append the last node's value to the result. For DFS: recursively traverse right subtree first, and if the current depth equals the result size, append the node's value.
State time complexity O(n) and space complexity O(n) for BFS (queue) or O(h) for DFS (recursion stack). Walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.