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Bytedance·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Apr 2026

Summary

Bytedance software engineer interview that went sideways fast once the coding portion started. Bombed the algorithm question badly enough that the interviewer just moved on.

Questions Asked (1)

Q1

Solve a tree-based algorithm problem involving finding the minimum number of increments to make a binary tree valid (LeetCode 2673 style).

Algorithms & Data Structures
Author's notes

Performed poorly and the interviewer just moved on.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem: given a binary tree where each node's value must equal the sum of its children's values (or 0 if leaf), find the minimum total increments to make it valid. Use a post-order DFS to compute the required value for each node based on its children, then sum the absolute differences between current and required values.

Pro tip: Emphasize that increments only increase values, so the required value for a node is the sum of its children's values (or 0 for leaves). This means the minimum increments at each node is simply the difference if the node's value is less than required; if greater, it's impossible, but the problem guarantees it's always possible by only incrementing leaves.

1. Clarify the problem and constraints

Confirm that the tree is binary, values are non-negative, and we can only increment node values. Ensure understanding that a valid tree requires each node's value to equal the sum of its children's values (0 for leaves).

2. Choose traversal and define recursive function

Use post-order DFS because children's values must be known before processing the parent. Define a function that returns the total increments needed in the subtree and the final value of the current node after increments.

3. Compute required value and increments

For a leaf, required value is 0. For an internal node, required value is the sum of its children's final values. If the node's current value is less than required, increment it by the difference and add that difference to the total increments. If greater, it's invalid, but the problem guarantees it won't happen.

4. Aggregate and return result

Recursively process left and right subtrees, sum their increments, then apply the current node's increment. Return the total increments for the entire tree.

5. Analyze complexity and edge cases

State that time complexity is O(n) since each node is visited once, and space complexity is O(h) for recursion stack. Mention edge cases like single node, skewed tree, and large values.

Key Points to Mention

  • Post-order traversal is necessary to process children before parent.
  • The required value for a node is the sum of its children's final values (0 for leaves).
  • Increments are only additive, so we only add the difference when current value is less than required.
  • The total increments is the sum of differences across all nodes.
  • Time complexity O(n) and space complexity O(h) due to recursion.
  • Edge cases: single node (0 increments), skewed tree (still O(n)), and large values (use 64-bit integers if needed).

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