Meta·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jun 2026Remote

Summary

Meta coding round, one tree problem the whole time. The question sounds straightforward but the O(1) space constraint is what actually makes it interesting, and I did not figure that out fast enough.

Questions Asked (1)

Given the root of a binary tree where each node has a 'next' pointer initialized to null, populate each node's next pointer to point to the node immediately to its right on the same level. Nodes with no right neighbor should have next set to null. Do this in O(n) time and O(1) extra space (no queue allowed). Return the root.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

My first instinct was BFS with a queue and I literally started coding it before they said no queue.

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AI HintsAI Generated

Suggested Approach

Use a level-order traversal without a queue by leveraging the already established next pointers of the previous level to traverse and link the current level. For each node, connect its left child to its right child, and if the node has a next, connect its right child to the next node's left child. This achieves O(n) time and O(1) space.

Pro tip: Emphasize that the O(1) space constraint is satisfied because we only use a few pointers and reuse the tree's structure; explicitly state that no recursion or queue is used, which is crucial for Meta's emphasis on optimal solutions.

1. Clarify the problem and constraints

Restate the problem to ensure understanding: populate next pointers for each node to point to its right neighbor on the same level, with O(n) time and O(1) extra space. Confirm that the tree is perfect (or handle general case) and that next pointers are initially null.

2. Outline the level-order traversal without a queue

Explain that you will process the tree level by level, using the next pointers of the current level to move to the next node. Start with the root as the leftmost node of the first level.

3. Describe the linking process for each node

For each node, if it has a left child, set left.next = right child. If the node has a next pointer, set right.next = node.next.left (if node.next exists). Then move to node.next to continue the level.

4. Move to the next level

After finishing the current level, move to the next level by setting the current node to the leftmost node of the next level, which is the left child of the first node of the current level (if it exists).

5. Analyze complexity and edge cases

State that time complexity is O(n) because each node is visited once, and space complexity is O(1) because only a constant number of pointers are used. Mention handling of edge cases like empty tree or single node.

Key Points to Mention

Leveraging existing next pointers to traverse levels without additional data structures.
The two-step linking: connecting children of the same parent, then connecting across parents using the parent's next pointer.
Time complexity O(n) and space complexity O(1) with justification.
Handling of edge cases: empty tree, single node, and incomplete levels (if not perfect tree).
Comparison with BFS using a queue, highlighting the space optimization.
Potential follow-up: general binary tree (not perfect) and how to adapt the approach.

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