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Apple·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Jun 2026

Summary

Apple SWE interview, got hit with the classic invert a binary tree problem. Pretty standard stuff but still a little nerve-wracking when it's Apple.

Questions Asked (1)

Q1

Invert a binary tree.

Algorithms & Data Structures
Author's notes

Knew this one cold from practice but still fumbled the explanation a bit.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and defining the binary tree node structure. Then present both recursive and iterative solutions, analyzing time and space complexity. Emphasize correctness, edge cases, and potential optimizations.

Pro tip: At Apple, interviewers value clean, efficient code and strong communication. Walk through your solution with a small example, and discuss trade-offs between recursion and iteration, especially regarding stack depth and memory usage.

1. Clarify the problem

Ask if the tree is binary, if nodes have parent pointers, and if the inversion should be done in-place. Confirm the definition of inversion (swap left and right children recursively).

2. Choose an approach

Decide between recursive and iterative (BFS/DFS) solutions. Recursive is simpler but may cause stack overflow for deep trees; iterative avoids recursion but uses extra space.

3. Implement the solution

Write clean code for your chosen approach. For recursion, swap children and recurse on left and right. For iteration, use a stack or queue to process nodes.

4. Test with examples

Walk through a small tree (e.g., 1 with left 2 and right 3) to verify correctness. Also consider edge cases: empty tree, single node, skewed tree.

5. Analyze complexity

State time complexity O(n) and space complexity O(h) for recursion (h = height) or O(n) for iterative with queue/stack. Discuss trade-offs.

Key Points to Mention

  • Definition of binary tree inversion: swapping left and right children at every node.
  • Recursive solution: base case (null node), swap children, recurse on left and right.
  • Iterative solution using stack (DFS) or queue (BFS) to avoid recursion overhead.
  • Time complexity: O(n) since each node is visited once.
  • Space complexity: O(h) for recursion (call stack) or O(n) for iterative (explicit data structure).
  • Edge cases: empty tree, single node, skewed tree, and potential stack overflow for deep trees.

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