I knew it was a level-order traversal thing pretty quickly, but I wasted maybe two minutes half-explaining DFS before catching myself.
Use a level-order traversal (BFS) with 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 O(n) time and O(h) space (where h is the height for DFS, or O(w) for BFS where w is max width).
Pro tip: Clarify with the interviewer whether the tree can be empty or have only one node, and discuss the trade-offs between BFS and DFS in terms of space complexity and implementation simplicity. Mention that BFS uses a queue and processes level by level, while DFS uses recursion (or stack) and can be more space-efficient for skewed trees.
Confirm that the right side view includes the rightmost node at each depth, even if it's not the right child of its parent. Clarify edge cases: empty tree, single node, skewed tree.
Decide between BFS (level-order traversal) and DFS (pre-order with right-first). BFS is intuitive: process each level and take the last node. DFS is elegant: track depth and record the first node seen at each depth when traversing right-first.
For BFS: use a queue, for each level, iterate through all nodes, and after the loop, add the last node's value to the result. For DFS: use recursion with a depth parameter, traverse right child first, and if depth equals result size, add node's value.
State time complexity O(n) since each node is visited once. Space complexity: BFS O(w) where w is maximum width, DFS O(h) where h is height (due to recursion stack).
Walk through a sample tree (e.g., [1,2,3,null,5,null,4]) to verify the output [1,3,4]. Also test edge cases: empty tree returns [], single node returns [value].
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.