I started with BFS since tracking depth per level feels natural there, but then they pushed on doing it in one pass for both views simultaneously.
Use a level-order traversal (BFS) with a queue, processing one level at a time. For each level, record the first node's value for the left view and the last node's value for the right view. This ensures both views are computed in a single pass with O(n) time and O(w) space, where w is the maximum width of the tree.
Pro tip: Mention that while BFS is natural for level-based views, a DFS with depth tracking can also work by updating the first and last seen nodes at each depth. However, BFS is more intuitive and avoids recursion stack overhead. Also, clarify that duplicate keys don't affect the view logic since we only care about node positions, not values.
Confirm that the tree is a BST (though the algorithm works for any binary tree) and discuss edge cases: empty tree returns empty lists, single node appears in both views, and duplicate keys are allowed but don't change the view logic.
Decide between iterative BFS (using a queue) and recursive DFS (with depth tracking). For a single-pass solution, BFS is straightforward: process level by level, capturing the first and last nodes.
Use a queue to traverse the tree. For each level, determine the number of nodes (level size). Iterate through the level, and for the first node, add its value to the left view; for the last node, add its value to the right view.
State that time complexity is O(n) since each node is visited once. Space complexity is O(w) for the queue, where w is the maximum width. Compare with DFS: O(h) space for recursion stack, but may require two passes or careful tracking.
Walk through a sample tree (e.g., a balanced BST) to verify both views. Also test edge cases like empty tree, skewed tree, and tree with duplicate keys to ensure correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.