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Meta·Software Engineer·Technical Phone Screen·Senior

Senior
Jun 2026

Summary

Meta Research Engineer coding round, got a tree problem that looks simple until you actually sit down and think about it under pressure.

Questions Asked (1)

Q1

Given the root of a binary tree where each node holds a single digit, every path from root to leaf forms a number. Write a function to return the sum of all such numbers.

Algorithms & Data Structures
Author's notes

I knew it was a DFS problem pretty quickly but fumbled the accumulation logic at first.

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AI HintsAI Generated

Suggested Approach

Use a depth-first search (DFS) to traverse the tree, maintaining the current number formed by the path from root to current node. At each leaf, add the current number to a running sum. This approach efficiently computes the sum in O(n) time and O(h) space, where h is the tree height.

Pro tip: Clarify with the interviewer whether the tree can be empty or contain negative digits, and discuss potential integer overflow if the tree is deep. Mention that you can use a 64-bit integer or modular arithmetic if needed.

1. Understand the problem

Confirm that each root-to-leaf path forms a number by concatenating digits, and the goal is to sum all such numbers. Ask about edge cases like empty tree, single node, or negative digits.

2. Choose traversal method

Decide between recursive DFS or iterative stack-based DFS. Recursive is simpler but may risk stack overflow for very deep trees; iterative avoids that but is more complex.

3. Design the recursive function

Define a helper function that takes a node and the current number formed so far. At each node, update the current number as current * 10 + node.val. If it's a leaf, return the current number; otherwise, return the sum of left and right subtrees.

4. Handle base cases and edge cases

If the root is null, return 0. If a node has only one child, continue the path without adding a zero. Ensure the algorithm works for negative digits if allowed.

5. Analyze complexity and test

State that time complexity is O(n) since each node is visited once, and space complexity is O(h) for recursion stack. Walk through a small example to verify correctness.

Key Points to Mention

  • Depth-first search (DFS) traversal
  • Maintaining the current number as you traverse
  • Handling leaf nodes by adding to sum
  • Time complexity O(n) and space complexity O(h)
  • Edge cases: empty tree, single node, negative digits
  • Potential integer overflow and mitigation strategies

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