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Meta·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Meta SWE coding round with a tree traversal problem. Pretty standard backtracking stuff but the negative numbers constraint is what trips people up if they're not careful.

Questions Asked (1)

Q1

Given a binary tree and a target sum, find all root-to-leaf paths where the sum of node values along the path equals the target. Return each valid path as a list of node values. The tree can contain negative numbers and up to 5000 nodes.

Algorithms & Data Structures
Author's notes

The negative numbers part is the thing I almost missed.

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AI HintsAI Generated

Suggested Approach

Use DFS with backtracking to explore all root-to-leaf paths, maintaining the current path and remaining sum. When a leaf is reached, check if the remaining sum equals the leaf's value; if so, add a copy of the current path to the result. Backtrack by removing the current node before returning.

Pro tip: Mention that you copy the path when adding to results to avoid mutation issues, and note that the problem guarantees at most 5000 nodes, so recursion depth is safe. Also, clarify that paths must end at a leaf, not just any node.

1. Clarify the problem

Confirm that a valid path must start at the root and end at a leaf, and that node values can be negative. Ask if the tree is binary and if the result should include paths as lists of integers.

2. Choose DFS with backtracking

Explain that DFS is ideal for exploring all root-to-leaf paths. Use a recursive function that carries the current path and the remaining target sum.

3. Implement the recursion

At each node, add its value to the path and subtract it from the remaining sum. If the node is a leaf and the remaining sum is zero, add a copy of the path to the result. Otherwise, recurse on left and right children.

4. Backtrack

After exploring both children, remove the current node from the path to restore the state for other branches. This ensures the path list is correctly maintained.

5. Analyze complexity

State that time complexity is O(N^2) in the worst case (e.g., a skewed tree) due to copying paths, but O(N) if we ignore copying. Space complexity is O(N) for the recursion stack and path storage.

Key Points to Mention

  • DFS with backtracking is the natural approach for path problems in trees.
  • Must check for leaf nodes (no children) before validating the sum.
  • Copy the current path when adding to results to avoid aliasing issues.
  • Negative numbers mean we cannot prune based on sum alone.
  • Time complexity: O(N^2) worst-case due to path copying, but often O(N) in practice.
  • Space complexity: O(N) for recursion stack and path storage.

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