← Bytedance Interview Insights

Bytedance·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Apr 2026Remote

Summary

Bytedance SWE interview with a binary tree problem that looked straightforward until the follow-up hit. One round, pretty algorithm-heavy, left me second-guessing my path reconstruction logic the whole way home.

Questions Asked (2)

Q1

Given the root of a binary tree, find the maximum path sum. A path can start and end at any node, doesn't need to go through the root, and each node can only be used once. Solve it in O(n) time.

Algorithms & Data Structures
Author's notes

Post-order DFS was the right call and I got there, but explaining why I needed to track the downward gain separately from the global max took me longer than it should have.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Use a post-order DFS to compute the maximum path sum. At each node, calculate the maximum gain from its left and right subtrees, update the global maximum with the sum of the node's value plus both gains, and return the node's value plus the larger gain to its parent.

Pro tip: Emphasize that the global maximum must be updated at every node, not just at the root, and discuss how to handle negative values by taking max(0, gain) to avoid including negative paths.

1. Clarify the problem

Confirm that a path can start and end at any node, must follow parent-child connections, and each node is used at most once. Ask if the tree can be empty or contain negative values.

2. Define recursive function

Define a helper function that returns the maximum gain (sum) from a subtree rooted at a given node, considering only paths that go straight down. This gain is the node's value plus the maximum of 0, left gain, and right gain.

3. Update global maximum

At each node, compute the sum of the node's value plus the left gain and right gain (if positive). Update the global maximum if this sum is larger.

4. Handle base cases and negative values

If the node is null, return 0. When computing gains, use max(0, gain) to ignore negative contributions, ensuring the path sum is maximized.

5. Analyze complexity

Explain that each node is visited once, so time complexity is O(n). Space complexity is O(h) for recursion stack, where h is tree height.

Key Points to Mention

  • Post-order traversal (DFS) to process children before parent
  • Global variable to track the maximum path sum found so far
  • At each node, consider the path that goes through the node and both children
  • Return value to parent should be the maximum gain from one side (node value + max(left gain, right gain))
  • Use max(0, gain) to avoid negative paths
  • Time complexity O(n), space complexity O(h) due to recursion

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

Q2

Follow-up: can you reconstruct and print the actual sequence of node values along that optimal path, not just the sum?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

This is where I fumbled.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Explain that you would augment the DP state to store the predecessor (or the action taken) that led to the optimal value at each state, then backtrack from the final state to reconstruct the path. Emphasize that this adds O(1) extra space per state and O(path length) time for reconstruction, without changing the overall time complexity.

Pro tip: Mention that you can often avoid storing full paths in each state by storing only the decision (e.g., which neighbor gave the best value) and then reconstructing the path by following those decisions backward. This keeps memory usage linear in the number of states, which is crucial for large inputs.

1. Clarify the DP formulation

Restate the DP recurrence and the meaning of each state, ensuring that the optimal value is computed correctly. Confirm that the DP table stores the best sum up to each state.

2. Augment state with predecessor info

For each state, store either the predecessor state or the action (e.g., direction) that yielded the optimal value. This can be a separate array or integrated into the DP table.

3. Backtrack to reconstruct the path

Starting from the final optimal state, follow the stored predecessor pointers backward until reaching the start state. Collect the node values along the way.

4. Reverse and output the path

Since backtracking yields the path in reverse order, reverse it to get the correct sequence from start to end. Print or return the sequence of node values.

5. Analyze complexity and trade-offs

Discuss the additional space (O(number of states)) and time (O(path length)) for reconstruction, and compare with alternative approaches like storing full paths in each state.

Key Points to Mention

  • DP state augmentation: storing predecessor or action per state
  • Backtracking from the optimal final state to the start
  • Time and space complexity: O(states) extra space, O(path length) reconstruction time
  • Trade-off between storing full paths vs. predecessor pointers
  • Handling multiple optimal paths (e.g., tie-breaking)
  • Edge cases: no path exists, single-node path, cycles (if applicable)

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