Meta·Software Engineer·Technical Phone Screen·Senior

Senior

May 2026

Summary

Meta coding round, one problem the whole session. The question looked like a tree traversal thing but had enough wrinkles that I spent a good chunk of time just making sure I understood what they were actually asking before writing a single line.

Questions Asked (1)

Given a binary tree, return two sequences: (1) a left-side view ordered from the deepest level up to the root, and (2) a right-side view ordered from the root down to the deepest level. Each view captures the first node visible at each depth from that side. Solve it in O(n) time with O(h) auxiliary space, and be ready to explain both a DFS and a BFS approach.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The two-views-in-one-problem thing tripped me up at first because I kept conflating them.

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AI HintsAI Generated

Suggested Approach

First, clarify the problem and edge cases, then present both BFS and DFS solutions, emphasizing the required O(n) time and O(h) space. For BFS, use a queue and record the first and last nodes at each level; for DFS, use recursion with depth tracking to update the first node seen at each depth from left and right. Finally, discuss trade-offs and how to reverse the left-side view to meet the ordering requirement.

Pro tip: Mention that the left-side view from deepest to root can be obtained by collecting the leftmost nodes during a top-down traversal and then reversing the list, which is O(h) extra space. This shows you understand both the traversal and the output ordering constraints.

1. Clarify the problem and constraints

Restate the problem: for each depth, the left-side view is the leftmost node, and the right-side view is the rightmost node. Confirm that the left-side view should be ordered from deepest level up to the root, and the right-side view from root down to the deepest level. Discuss edge cases: empty tree, single node, skewed tree.

2. Present the BFS approach

Use a queue to perform level-order traversal. For each level, the first node polled is the left-side view, and the last node polled is the right-side view. Collect these nodes in lists, then reverse the left-side list to get the deepest-to-root order. Analyze time O(n) and space O(w) where w is max width; note that O(w) can be O(n) in worst case, so it does not meet the O(h) auxiliary space requirement.

3. Present the DFS approach

Use recursion (or an explicit stack) to traverse the tree, passing the current depth. Maintain two arrays (or maps) to store the first node encountered at each depth from the left and from the right. For left-side view, update when visiting a depth for the first time in a pre-order traversal (node, left, right). For right-side view, update when visiting a depth for the first time in a reverse pre-order traversal (node, right, left). After traversal, the left-side array is in root-to-deepest order; reverse it to get deepest-to-root. The right-side array is already in root-to-deepest order. Space is O(h) for recursion stack plus O(h) for the arrays, which meets the requirement.

4. Compare and contrast the approaches

Discuss the trade-offs: BFS is intuitive and directly gives level-order views but uses O(w) space, which can be O(n) in the worst case. DFS uses O(h) space, which is optimal for balanced trees, but requires careful ordering and reversing. Mention that both are O(n) time. Highlight that the problem explicitly asks for O(h) auxiliary space, so DFS is preferred.

5. Discuss extensions and edge cases

Mention how to handle very deep trees (recursion depth) by using an iterative DFS with an explicit stack. Discuss how to adapt if the tree is not binary or if multiple nodes are visible from the side (e.g., if nodes can overlap). Also, consider if the tree is modified during traversal (not typical).

Key Points to Mention

Time complexity O(n) for both BFS and DFS because each node is visited once.
Space complexity: BFS uses O(w) where w is maximum width, which can be O(n); DFS uses O(h) for recursion stack and result storage, meeting the O(h) requirement.
For DFS, use pre-order (node-left-right) to capture leftmost nodes and reverse pre-order (node-right-left) to capture rightmost nodes.
The left-side view needs to be reversed to match the deepest-to-root order; the right-side view is naturally root-to-deepest.
Edge cases: empty tree returns empty lists; single node returns [node] for both views; skewed tree (all left or all right) results in views of length equal to tree height.
Trade-offs: BFS is simpler for level-order but uses more space; DFS is more space-efficient but requires careful ordering and possibly reversing.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.