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Google·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Jun 2026

Summary

Google SWE coding round, one question the whole time. Pretty standard tree problem but I still managed to second-guess myself halfway through.

Questions Asked (1)

Q1

Given a binary tree, implement a function that traverses it level by level from left to right and returns the node values grouped by level.

Algorithms & Data Structures
Author's notes

BFS with a queue, pretty textbook.

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AI HintsAI Generated

Suggested Approach

Clarify the problem and edge cases, then propose a BFS solution using a queue to process nodes level by level. Explain how to track level boundaries and collect values into sublists, and analyze time and space complexity.

Pro tip: Mention that BFS is ideal for level-order traversal, but also discuss how DFS with level tracking can achieve the same result, showing versatility. Emphasize the importance of handling edge cases like an empty tree and skewed trees.

1. Clarify requirements and edge cases

Confirm the output format (list of lists), input constraints, and edge cases such as empty tree, single node, and skewed trees.

2. Choose BFS with a queue

Explain that BFS naturally processes nodes level by level. Use a queue to store nodes and process each level by iterating over the current queue size.

3. Implement level tracking

For each level, record the number of nodes (queue size), dequeue that many nodes, collect their values, and enqueue their children.

4. Analyze complexity

State that time complexity is O(N) where N is the number of nodes, and space complexity is O(W) where W is the maximum width of the tree.

5. Discuss alternatives and optimizations

Mention DFS with level parameter as an alternative, and discuss potential optimizations like using a deque or handling large trees.

Key Points to Mention

  • BFS uses a queue (FIFO) to process nodes level by level.
  • Track level boundaries by recording the queue size at the start of each level.
  • Time complexity: O(N) since each node is visited once.
  • Space complexity: O(W) where W is the maximum width of the tree.
  • Edge cases: empty tree returns empty list, single node returns [[value]].
  • Alternative: DFS with level tracking can also solve this, but BFS is more intuitive.

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