Knew this one the second they said 'binary tree' and 'visible nodes'.
Clarify whether the view is from the right or left side, then perform a level-order traversal (BFS) and record the last node's value at each level for the right view (or first node for the left view). Alternatively, use DFS with depth tracking, updating the result when visiting a node at a new depth, prioritizing the appropriate child order.
Pro tip: Mention that the problem is essentially finding the rightmost (or leftmost) node at each depth, and that BFS is often more intuitive but DFS uses less memory for skewed trees. Also, discuss handling edge cases like an empty tree and the importance of clarifying the view direction upfront.
Ask whether the view is from the right or left side, and confirm the expected output format (list of values top to bottom).
Decide between BFS (level-order traversal) and DFS (depth-first traversal with depth tracking). Explain the trade-offs.
For BFS, use a queue to process nodes level by level, capturing the last (or first) node's value. For DFS, recursively traverse, prioritizing the right (or left) child, and record the first node seen at each depth.
Consider an empty tree, a tree with only one node, and skewed trees. Ensure the solution works for these cases.
State that both approaches have O(n) time complexity and O(n) space in the worst case (for BFS, the queue; for DFS, the recursion stack).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.