Meta·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Meta SWE interview, tree problem with a tricky space constraint. Not the hardest question on paper but the O(1) space requirement is where it gets interesting.

Questions Asked (1)

Given a binary tree where each node has a `next` pointer initialized to NULL, populate every node's `next` to point to the next node at the same level (or leave NULL if none exists). You must do this in O(1) extra space by leveraging the `next` chain you've already built on the parent level to traverse and wire up the children.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The basic idea clicked pretty fast, BFS-style level linking.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and constraints, then propose a level-order traversal that uses the already established next pointers to avoid extra space. Explain the algorithm step-by-step, handling both left and right children, and analyze time and space complexity.

Pro tip: Emphasize that the O(1) space is achieved by reusing the next pointers as a linked list for the next level, and mention that this approach works even for perfect binary trees but also handles incomplete trees with careful checks.

1. Clarify the problem and constraints

Confirm that the tree may not be perfect, that next pointers are initially NULL, and that O(1) extra space means no queues or recursion stack.

2. Outline the level-order traversal using next pointers

Explain that you will traverse each level using the next pointers already set on the parent level, and while traversing, set the next pointers for the children.

3. Detail the child connection logic

For each node, if it has a left child, set its next to the right child; if it has a right child, set its next to the left child of the node's next (if exists).

4. Handle incomplete trees and edge cases

Mention that you need to check for null children and that the next pointer of a node may be null, so you must skip to the next available node.

5. Analyze complexity and discuss trade-offs

State that time complexity is O(n) and space is O(1), and discuss why this is optimal and any potential pitfalls.

Key Points to Mention

Use the next pointers of the current level to traverse the next level without additional data structures.
Maintain a pointer to the first node of the next level to start the next iteration.
When connecting children, consider both left and right children and the case where the parent's next is null.
Time complexity is O(n) because each node is visited once.
Space complexity is O(1) because only a few pointers are used.
The algorithm works for any binary tree, not just perfect ones, with proper null checks.

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