Microsoft·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Microsoft SWE interview with a tree problem that looks straightforward until you hit the space constraint. The O(1) extra space requirement is what separates people who've seen this before from everyone else.

Questions Asked (1)

Given a perfect binary tree where every node has a 'next' pointer, connect each node's next pointer to its right neighbor at the same level. Nodes with no right neighbor should point to null. Solve it using O(1) extra space.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

My first instinct was BFS with a queue and I even started explaining it before they stopped me and pointed at the space requirement.

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AI HintsAI Generated

Suggested Approach

Use the already-established next pointers at each level to traverse the next level without using a queue, achieving O(1) space. For each node, connect its left child's next to its right child, and if the node has a next, connect its right child's next to the left child of its next node. Iterate level by level until all levels are processed.

Pro tip: Explicitly state that the O(1) space constraint rules out BFS with a queue, and emphasize that you're leveraging the next pointers as a linked list to move horizontally. This shows you understand the trade-off and can optimize space.

1. Clarify the problem and constraints

Confirm that the tree is perfect (all leaves at same level, each internal node has two children) and that the next pointer is initially null. Restate the O(1) space requirement to ensure alignment.

2. Identify the key insight

Recognize that once a level is connected, its next pointers form a linked list, allowing traversal of the next level without extra space. This enables a level-by-level connection using only a few pointers.

3. Design the algorithm

Start at the root. For each level, use a pointer to traverse nodes via next. For each node, set left.next = right, and if node.next exists, set right.next = node.next.left. Move to the next level by setting current to the leftmost node of the next level.

4. Analyze complexity and edge cases

Time complexity is O(n) since each node is visited once. Space is O(1) as only a constant number of pointers are used. Handle edge cases like empty tree (return null) and single node (next remains null).

5. Test with a small example

Walk through a perfect binary tree of height 3 (7 nodes) to verify connections. Check that all next pointers are correctly set and that the last node on each level points to null.

Key Points to Mention

The O(1) space constraint eliminates queue-based BFS, so we must use the next pointers themselves to traverse levels.
The algorithm processes one level at a time, using the already-connected next pointers of the current level to connect the next 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 left child of its next node.
Time complexity is O(n) because each node is visited once; space complexity is O(1) due to constant extra pointers.
Edge cases: empty tree returns null; a tree with only the root has its next set to null.
The solution works only for perfect binary trees; for general trees, a different approach or extra space might be needed.

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