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Apple·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jul 2026

Summary

Apple SWE interview with a tree traversal problem. Nothing too wild but the tie-breaking rule for same-row same-column nodes is the kind of detail that will absolutely trip you up if you're not careful.

Questions Asked (1)

Q1

Given the root of a binary tree, return the vertical order traversal of its node values, grouped by column from leftmost to rightmost. Within each column, nodes should be ordered top to bottom by depth, and if two nodes share the same row and column, list them left to right in BFS order.

Algorithms & Data Structures
Author's notes

The basic column-tracking part I got pretty fast, just assign an index to each node and collect values by column.

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AI HintsAI Generated

Suggested Approach

Use BFS to traverse the tree level by level, tracking each node's column index (root at 0, left child -1, right child +1). Store nodes in a hash map keyed by column, appending values as encountered; then output columns from min to max. BFS naturally ensures top-to-bottom and left-to-right ordering within each column.

Pro tip: Clarify the tie-breaking rule upfront: if two nodes share the same row and column, BFS order ensures left-to-right. Mention that DFS with sorting could work but BFS is more efficient and directly satisfies the ordering constraints.

1. Clarify requirements and edge cases

Confirm the ordering rules: columns left to right, within column top to bottom, and same row/column left to right. Ask about empty tree, single node, and nodes with same column but different depths.

2. Choose BFS with column tracking

Use a queue for level-order traversal, storing each node with its column index. Initialize root at column 0; left child gets col-1, right child col+1.

3. Group nodes by column

Use a hash map (or dictionary) mapping column index to a list of node values. Append values as nodes are dequeued, which preserves top-to-bottom and left-to-right order.

4. Collect and return result

Determine the min and max column indices, then iterate from min to max, appending each column's list to the result. Return the list of lists.

Key Points to Mention

  • BFS ensures nodes are processed in order of depth (top to bottom) and left to right within the same level.
  • Column index tracking: root at 0, left child -1, right child +1.
  • Hash map (or dictionary) to group node values by column index.
  • Time complexity: O(n) where n is number of nodes; space complexity: O(n) for queue and map.
  • Edge cases: empty tree returns empty list; single node returns [[root.val]].
  • Alternative DFS approach would require sorting by depth and horizontal position, which is less efficient.

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