← Meta Interview Insights

Meta·Machine Learning Engineer·Technical Phone Screen·Senior

Senior
May 2026

Summary

Meta MLE interview with a tree traversal problem that sounds easy until you're actually in it. They wanted both iterative and recursive solutions plus a real discussion of tradeoffs, not just code that runs.

Questions Asked (1)

Q1

Given a binary tree, return the values visible from the left side when viewed top to bottom. At each depth level, only the leftmost node is visible. Implement both an iterative level-order traversal solution and a recursive solution, analyze time and space complexity for each, and handle edge cases like skewed trees and missing children.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

I went straight to the recursive solution first because it felt more natural, then had to backtrack and build the iterative BFS version.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and edge cases, then present both iterative BFS and recursive DFS solutions with clear code and complexity analysis. Emphasize the trade-offs between the two approaches and how they handle skewed trees and missing children.

Pro tip: In interviews, always discuss edge cases like empty tree, single node, and skewed trees before coding, and mention that the recursive solution can be optimized to O(1) space if recursion stack is not counted, but be transparent about the trade-offs.

1. Clarify and Define

Restate the problem to ensure understanding: left side view means the first node at each depth. Discuss edge cases: empty tree, single node, skewed left/right, and missing children.

2. Iterative BFS Solution

Explain level-order traversal using a queue. At each level, record the first node's value. Handle missing children by only enqueuing non-null nodes.

3. Recursive DFS Solution

Use pre-order traversal (root, left, right) and track the current depth. If the depth is visited for the first time, add the node's value to the result. This ensures the leftmost node at each depth is recorded.

4. Complexity Analysis

For both solutions, time complexity is O(n) where n is number of nodes. Space: BFS O(w) where w is max width; recursive O(h) where h is height (skewed tree O(n)).

5. Trade-offs and Edge Cases

Compare BFS vs DFS: BFS uses more memory for wide trees, DFS uses recursion stack. Discuss handling skewed trees and missing children in both implementations.

Key Points to Mention

  • Time complexity O(n) for both solutions, space complexity O(w) for BFS and O(h) for DFS.
  • Iterative BFS uses a queue and processes level by level, capturing the first node at each level.
  • Recursive DFS uses pre-order traversal and a depth parameter to capture the first node at each depth.
  • Edge cases: empty tree returns empty list, single node returns [root.val], skewed trees (left or right) affect space complexity.
  • Missing children: in BFS, only enqueue non-null children; in DFS, base case handles null nodes.
  • Trade-offs: BFS is intuitive for level-order but uses more memory for wide trees; DFS is memory-efficient for skewed trees but recursion depth may be a concern.

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