Meta·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jul 2026

Summary

Meta software engineer interview with a tree problem that sounds straightforward but has some gotchas if you're not careful with how you propagate subtree info.

Questions Asked (1)

Given a binary tree, check whether every node's value equals the integer average of all values in its own subtree (including itself).

Algorithms & Data Structures

Author's notes

My first instinct was to write a separate helper that recomputes the subtree sum and count from scratch for each node, which works but is O(n^2) and they pushed back on it pretty fast.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Use a post-order traversal to compute each subtree's sum and node count, then verify that the node's value equals the integer average (sum divided by count). Return both the sum and count (or a boolean flag) from each recursive call to avoid redundant traversals.

Pro tip: Clarify the definition of 'integer average' upfront—whether it's floor, round, or exact division—and handle edge cases like empty trees or single nodes. Also, mention that you can short-circuit and return early if any subtree fails the condition.

1. Clarify requirements and edge cases

Ask the interviewer to confirm the definition of 'integer average' (e.g., floor, round, or exact) and discuss edge cases such as an empty tree, a single node, or negative values.

2. Choose traversal and return values

Decide on a post-order traversal (left, right, root) because you need subtree information before checking the current node. Plan to return both the sum of values and the number of nodes in the subtree.

3. Design recursive function

Write a helper function that returns a pair (sum, count) and a boolean indicating whether the subtree satisfies the condition. At each node, combine left and right results, compute the average, and compare with the node's value.

4. Implement and test

Code the solution iteratively or recursively, ensuring integer division is handled correctly. Test with simple cases (single node, perfect tree) and edge cases (unbalanced tree, negative values).

5. Analyze complexity and optimize

State that the time complexity is O(n) since each node is visited once, and space complexity is O(h) for recursion stack. Mention that early termination can improve average-case performance.

Key Points to Mention

Post-order traversal is necessary because we need subtree sums and counts before checking the current node.
Returning both sum and count from each recursive call avoids multiple traversals and keeps time complexity O(n).
Integer division details: clarify whether to use floor, round, or exact division, and handle potential division by zero (though count is always ≥1).
Edge cases: empty tree (return true), single node (always true), and negative values (ensure average calculation works).
Space complexity: O(h) for recursion stack, where h is tree height; can be O(n) in worst case (skewed tree).
Early termination: if any subtree fails, propagate false immediately to avoid unnecessary checks.

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