Bytedance·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jun 2026

Summary

Bytedance SWE interview with a tree problem that looks straightforward until you hit the space constraint. Not a bad experience overall but the follow-up on constant space tripped me up.

Questions Asked (1)

Given a perfect binary tree where every parent has two children and all leaves are on the same level, populate each node's next pointer to point to its next right neighbor on the same level. If no right neighbor exists, set next to NULL. Solve it using only constant extra space.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

Got the BFS version out pretty fast, which felt good, then they asked me to redo it without a queue.

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AI HintsAI Generated

Suggested Approach

Use the already-established next pointers to traverse each level like a linked list, connecting children of adjacent nodes without any queue. Emphasize that the perfect binary tree property guarantees every node has both children, simplifying the connection logic. Explain the O(n) time and O(1) space complexity, and contrast with the BFS approach that uses O(n) space.

Pro tip: Mention that the perfect binary tree property is crucial for the O(1) space solution; if the tree were not perfect, you'd need additional checks or a different approach. Also, clarify that the next pointers themselves are used as the 'queue' for the next level, which is the key insight.

1. Clarify the problem and constraints

Restate that the tree is perfect (every parent has two children, all leaves at same level) and that we must use O(1) extra space. Confirm that modifying the tree in-place is allowed.

2. Explain the BFS approach and its space complexity

Describe how a level-order traversal with a queue would work, but note that it uses O(n) space, which violates the constant space requirement.

3. Introduce the O(1) space approach using next pointers

Explain that we can traverse each level using the next pointers already set for that level, and use that to set next pointers for the next level. Start with the root's level (only root), then move down.

4. Detail the connection logic for children

For each node in the current level, set its left child's next to its right child, and if the node has a next, set its right child's next to the next node's left child. Move to the next node in the current level via the next pointer.

5. Analyze complexity and edge cases

State that time complexity is O(n) since each node is visited once, and space is O(1) because we only use a few pointers. Mention edge cases: empty tree, single node, and that the perfect property ensures no missing children.

Key Points to Mention

The perfect binary tree property guarantees every internal node has both left and right children, so no null checks for children are needed.
Using the next pointers of the current level as a linked list to traverse it, which avoids using a queue.
The two connection rules: left child's next = right child; right child's next = (current node's next)?.left.
Time complexity O(n) and space complexity O(1), which meets the constant space requirement.
Comparison with BFS approach that uses O(n) space, highlighting the trade-off.
The solution modifies the tree in-place and does not require any additional data structures.

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