Uber·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Apr 2026

Summary

Uber SWE interview with a tree manipulation problem. Pretty standard algorithmic round but the O(1) space constraint is what makes it actually interesting.

Questions Asked (1)

Given a perfect binary tree where every leaf is at the same depth, populate each node's 'next' pointer to point to its next right neighbor on the same level, or NULL if none exists. Try to do it with O(1) extra space.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The naive BFS solution comes to mind immediately but they pushed back on the space usage right away.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and constraints, then propose a level-by-level traversal using already established next pointers to avoid queues or recursion. Emphasize the O(1) space requirement by leveraging the tree structure and parent-level connections.

Pro tip: Mention that this approach works because the tree is perfect; for a general binary tree, O(1) space is not possible without parent pointers or additional data structures. This shows you understand the problem's boundaries.

1. Clarify the problem and constraints

Confirm that the tree is perfect, that we need to set next pointers for all nodes, and that O(1) extra space means no queues or recursion stack.

2. Outline the level-order traversal idea

Explain that we can traverse each level using the next pointers already set for the parent level, starting from the leftmost node of each level.

3. Detail the connection logic

For each node, 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.

4. Handle level transitions

After processing a level, move to the next level by following the left child of the leftmost node, and repeat until all levels are processed.

5. Analyze complexity and edge cases

State that time complexity is O(n) and space is O(1). Discuss edge cases like empty tree or single node.

Key Points to Mention

Perfect binary tree property ensures every node except leaves has two children, simplifying connections.
Using next pointers of the current level to traverse the next level without additional data structures.
The two key assignments: left.next = right, and right.next = (current.next != null) ? current.next.left : null.
Time complexity O(n) because each node is visited once; space O(1) because only a few pointers are used.
Comparison with BFS using a queue (O(n) space) to highlight the optimization.
Potential follow-up: how to adapt for a general binary tree (e.g., using a dummy node for each level).

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