Went with BFS since level-order traversal makes it pretty natural to grab the last node at each depth.
Clarify the problem and edge cases, then present a level-order BFS solution that records the last node at each level. Explain the algorithm step-by-step, analyze time and space complexity, and briefly mention an alternative DFS approach for comparison.
Pro tip: Emphasize that BFS naturally processes nodes level by level, making it intuitive to capture the rightmost node; also note that DFS can achieve O(h) space, which is advantageous for deep trees.
Confirm that 'right side view' means the rightmost node at each depth, and discuss edge cases like an empty tree or a tree with only left children.
Select BFS for its intuitive level-order traversal, or DFS if space efficiency is critical; briefly justify your choice.
For BFS: use a queue, process each level, and record the last node's value. For DFS: traverse right-first, tracking depth and updating the result when visiting a new depth.
State that both approaches run in O(n) time. BFS uses O(w) space (w = max width), while DFS uses O(h) space (h = height).
Compare BFS and DFS in terms of space usage, code simplicity, and suitability for different tree shapes.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.