Meta·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Apr 2026

Summary

Meta SWE interview with a tree problem that looks clean on the surface but has enough edge cases to trip you up if you haven't thought carefully about tie-breaking rules.

Questions Asked (1)

Given the root of a binary tree, group nodes into vertical columns by assigning each node an x-coordinate (root = 0, left child = x-1, right child = x+1). For each column from leftmost to rightmost, list node values top to bottom. When multiple nodes share the same x-coordinate and depth, order them by their left-to-right level-order position. Return all columns as a list of lists. Walk through your traversal approach, how you handle ties, and what data structures you'd use. Give time and space complexity.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The x-coordinate assignment part I got down fast.

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AI HintsAI Generated

Suggested Approach

Use a BFS level-order traversal while tracking each node's x-coordinate, storing nodes in a map from x-coordinate to a list of values. After traversal, sort the x-coordinates and output the lists in order. For tie-breaking, process nodes level by level from left to right, so nodes at the same depth and x-coordinate are naturally ordered by their left-to-right position.

Pro tip: Clarify the tie-breaking rule upfront: if two nodes share the same x-coordinate and depth, the one encountered first in a left-to-right level-order traversal comes first. This shows attention to detail and avoids ambiguity.

1. Choose traversal

Use BFS (level-order) to ensure top-to-bottom and left-to-right ordering. Alternatively, DFS with depth tracking can work but requires sorting by depth and position.

2. Track coordinates

Assign x-coordinate to each node: root at 0, left child x-1, right child x+1. Store nodes in a hash map keyed by x-coordinate, with values as lists of node values.

3. Handle ties

During BFS, process nodes level by level from left to right. This ensures that for the same x-coordinate and depth, nodes are added in left-to-right order.

4. Collect and sort columns

After traversal, extract all x-coordinates, sort them, and for each, output the corresponding list of values.

5. Analyze complexity

Time: O(N log N) due to sorting x-coordinates (or O(N) if using ordered map). Space: O(N) for the map and queue.

Key Points to Mention

BFS ensures top-to-bottom and left-to-right ordering within each column.
Use a hash map (or ordered map) to group nodes by x-coordinate.
Tie-breaking: nodes at same x and depth are ordered by their left-to-right position in level-order.
Time complexity: O(N log N) with sorting, or O(N) with ordered map; space O(N).
Edge cases: empty tree, skewed tree, nodes with same x but different depths.
Alternative: DFS with (x, depth) and sorting by depth and position, but BFS is simpler.

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