Vertical Order Traversal of a Binary Tree

Vertical Order Traversal of a Binary Tree

What is this problem about?

The "Vertical Order Traversal of a Binary Tree" interview question is a "Hard" tree problem that asks you to return the values of nodes grouped by their vertical columns. For each node at (row, col), its left child is at (row + 1, col - 1) and its right child is at (row + 1, col + 1). If multiple nodes share the same row and col, they must be sorted by value.

Why is this asked in interviews?

Tech giants like Microsoft, Meta, and Google use this problem to test a candidate's mastery of tree traversals (BFS/DFS) and their ability to organize data using multi-level sorting. It’s a "Binary Tree interview pattern" problem that requires managing coordinates and complex nested data structures.

Algorithmic pattern used

The primary pattern is "Breadth-First Search (BFS)" or "Depth-First Search (DFS)" combined with a "Sorting" strategy. You traverse the tree and store each node as a triplet: (column, row, value). After collecting all nodes, you sort them primarily by column, secondarily by row, and tertiarily by value. Finally, you group the sorted nodes by column for the final output.

Example explanation

Tree: 1 at root, 2 on left, 3 on right.

1. Root (1): (col:0, row:0, val:1).
2. Left (2): (col:-1, row:1, val:2).
3. Right (3): (col:1, row:1, val:3).
4. Nodes: {(-1,1,2), (0,0,1), (1,1,3)}.
5. Sorted by column: Column -1: [2], Column 0: [1], Column 1: [3].
6. Result: [[2], [1], [3]].

Common mistakes candidates make

One common mistake is using a simple BFS that only groups by column, forgetting the requirement to sort by value when nodes have the same row and column. Another error is not correctly managing the range of column indices (which can be negative), which is why a TreeMap or sorting the final keys is often necessary.

Interview preparation tip

For the "Vertical Order Traversal of a Binary Tree" coding problem, BFS is often preferred over DFS because it naturally visits nodes in row order. This simplifies the sorting requirements, as you only need to worry about nodes at the exact same row and column.