Vertical Order Traversal of Binary Tree in DSA Javascript

Mental Model

We want to group tree nodes by their vertical columns from left to right, and list nodes top to bottom in each column.

Analogy: Imagine standing in front of a tree and looking straight at it. Nodes that line up vertically in your view belong to the same group. We collect these groups from left side to right side.

      1
     / \
    2   3
   / \   \
  4   5   6

Columns: -2  -1   0   1   2
Nodes:   4   2   1,5  3   6

Dry Run Walkthrough

Input: Binary tree: 1 / \ 2 3 / \ \ 4 5 6

Goal: Group nodes by vertical columns from left to right, top to bottom in each column

Step 1: Start at root node 1 with column 0

Column 0: [1]
Columns so far: {0: [1]}

Why: Root is the center column, start grouping here

Step 2: Move to left child 2 with column -1

Column -1: [2]
Columns so far: {-1: [2], 0: [1]}

Why: Left child is one column to the left

Step 3: Move to left child 4 with column -2

Column -2: [4]
Columns so far: {-2: [4], -1: [2], 0: [1]}

Why: Left child again moves one column left

Step 4: Back to node 2, move to right child 5 with column 0

Column 0: [1, 5]
Columns so far: {-2: [4], -1: [2], 0: [1, 5]}

Why: Right child moves one column right from parent

Step 5: Back to root 1, move to right child 3 with column 1

Column 1: [3]
Columns so far: {-2: [4], -1: [2], 0: [1, 5], 1: [3]}

Why: Right child moves one column right

Step 6: Move to right child 6 of node 3 with column 2

Column 2: [6]
Columns so far: {-2: [4], -1: [2], 0: [1, 5], 1: [3], 2: [6]}

Why: Right child moves one column right

Result:

Columns from left to right:
-2: 4
-1: 2
0: 1 -> 5
1: 3
2: 6

Annotated Code

DSA Javascript

class TreeNode {
  constructor(val, left = null, right = null) {
    this.val = val;
    this.left = left;
    this.right = right;
  }
}

function verticalOrderTraversal(root) {
  if (!root) return [];
  const columnTable = new Map();
  const queue = [];
  queue.push([root, 0]); // node with its column index

  while (queue.length > 0) {
    const [node, col] = queue.shift();
    if (!columnTable.has(col)) {
      columnTable.set(col, []);
    }
    columnTable.get(col).push(node.val); // add node value to its column

    if (node.left) {
      queue.push([node.left, col - 1]); // left child column is col-1
    }
    if (node.right) {
      queue.push([node.right, col + 1]); // right child column is col+1
    }
  }

  const sortedCols = Array.from(columnTable.keys()).sort((a, b) => a - b);
  const result = [];
  for (const col of sortedCols) {
    result.push(columnTable.get(col));
  }
  return result;
}

// Driver code
const root = new TreeNode(1,
  new TreeNode(2,
    new TreeNode(4),
    new TreeNode(5)
  ),
  new TreeNode(3,
    null,
    new TreeNode(6)
  )
);

const output = verticalOrderTraversal(root);
for (const colNodes of output) {
  console.log(colNodes.join(' -> '));
}

queue.push([root, 0]); // node with its column index

Initialize queue with root node at column 0

const [node, col] = queue.shift();

Take next node and its column from queue for processing

if (!columnTable.has(col)) { columnTable.set(col, []); }

Create list for column if not exists

columnTable.get(col).push(node.val); // add node value to its column

Add current node value to its column group

if (node.left) { queue.push([node.left, col - 1]); }

Add left child to queue with column one less

if (node.right) { queue.push([node.right, col + 1]); }

Add right child to queue with column one more

const sortedCols = Array.from(columnTable.keys()).sort((a, b) => a - b);

Sort columns from left to right

for (const col of sortedCols) { result.push(columnTable.get(col)); }

Collect node groups in column order

Output

4

2

1 -> 5

3

6

Complexity Analysis

Time: O(n) because we visit each node once during BFS traversal

Space: O(n) because we store all nodes in the column table and queue

vs Alternative: Compared to DFS, BFS preserves top-to-bottom order naturally without extra sorting by depth

Edge Cases

Empty tree (root is null)

Returns empty list immediately

DSA Javascript

if (!root) return [];

Tree with only one node

Returns list with one column containing that single node

DSA Javascript

queue.push([root, 0]);

Tree with multiple nodes in same column

Nodes appear in order top to bottom as visited by BFS

DSA Javascript

columnTable.get(col).push(node.val);

When to Use This Pattern

When you need to group tree nodes by vertical columns and preserve top-to-bottom order, use vertical order traversal with BFS and column indexing.

Common Mistakes

Mistake: Using DFS without tracking node depth causes wrong order in columns
Fix: Use BFS to process nodes level by level to maintain top-to-bottom order

Mistake: Not sorting columns before output leads to unordered columns
Fix: Sort column keys before collecting results

Summary

Groups binary tree nodes by their vertical columns from left to right, top to bottom.

Use when you want to see nodes aligned vertically as if viewing the tree from the front.

The key is to assign column indices and use BFS to keep nodes in top-down order within each column.

Practice

1. What does Vertical Order Traversal of a binary tree mean?

easy

A. Nodes are visited level by level from top to bottom.

B. Nodes are visited in preorder (root, left, right).

C. Nodes are grouped by their horizontal rows from top to bottom.

D. Nodes are grouped by their vertical columns from left to right.

Solution

1. Step 1: Understand vertical columns in a binary tree

   Vertical order groups nodes by their horizontal distance from the root, called columns.

2. Step 2: Identify traversal order

   Nodes in the same vertical column are collected and output from leftmost to rightmost column.

3. Final Answer:

   Nodes are grouped by their vertical columns from left to right. -> Option D

4. Quick Check:

   Vertical order = group by columns [OK]

Hint: Think columns, not rows or preorder traversal [OK]

Common Mistakes:

• Confusing vertical order with level order traversal
• Thinking vertical order is preorder or inorder traversal
• Mixing horizontal rows with vertical columns

2. Which of the following JavaScript code snippets correctly initializes a map to store vertical order nodes by column?

easy

A. const map = new Map();

B. const map = {};

C. const map = new Set();

D. const map = [];

Solution

1. Step 1: Identify suitable data structure for key-value pairs

   We need to map column numbers (keys) to arrays of nodes (values), so Map is ideal.

2. Step 2: Check each option

   const map = new Map(); uses Map which supports keys of any type and preserves insertion order. const map = {}; is an object but less flexible for numeric keys. const map = new Set(); is a Set, which stores unique values without keys. const map = []; is an array, not suitable for key-value mapping.

3. Final Answer:

   const map = new Map(); -> Option A

4. Quick Check:

   Use Map for column to nodes mapping [OK]

Hint: Use Map for flexible key-value storage [OK]

Common Mistakes:

• Using Set instead of Map for key-value pairs
• Using array instead of Map for mapping columns
• Using plain object which may cause key type issues

3. Given the binary tree:

    1
   / \
  2   3
 / \   \
4   5   6

What is the vertical order traversal output?

medium

A. [[4], [2], [1, 5], [3], [6]]

B. [[1], [2, 3], [4, 5, 6]]

C. [[4, 2], [1, 5, 3], [6]]

D. [[4], [2, 5], [1, 3, 6]]

Solution

1. Step 1: Assign column indices to nodes

   Root (1) is column 0. Left child decreases column by 1, right child increases by 1.
   So, 4 is at -2, 2 at -1, 1 and 5 at 0, 3 at 1, 6 at 2.

2. Step 2: Group nodes by columns and list them top to bottom

   Columns: -2: [4], -1: [2], 0: [1,5], 1: [3], 2: [6]

3. Final Answer:

   [[4], [2], [1, 5], [3], [6]] -> Option A

4. Quick Check:

   Column grouping matches answer A [OK]

Hint: Track column index: left -1, right +1 [OK]

Common Mistakes:

• Mixing node order within the same column
• Ignoring column indices and grouping by levels
• Swapping left and right child column adjustments

4. Consider this code snippet for vertical order traversal:

function verticalOrder(root) {
  const map = new Map();
  function dfs(node, col) {
    if (!node) return;
    dfs(node.left, col - 1);
    dfs(node.right, col + 1);
    if (!map.has(col)) map.set(col, []);
    map.get(col).push(node.val);
  }
  dfs(root, 0);
  return [...map.values()];
}

What is the main issue with this code?

medium

A. The function does not handle empty trees.

B. Map keys should be strings, not numbers.

C. DFS does not guarantee top-to-bottom order in vertical columns.

D. The code uses recursion incorrectly.

Solution

1. Step 1: Understand DFS traversal order

   DFS explores depth first, so nodes deeper in the tree may be visited before nodes higher up in the same column.

2. Step 2: Identify vertical order requirement

   Vertical order requires nodes in each column to be output top to bottom, which DFS alone does not guarantee.

3. Final Answer:

   DFS does not guarantee top-to-bottom order in vertical columns. -> Option C

4. Quick Check:

   DFS order ≠ vertical top-to-bottom order [OK]

Hint: Use BFS for correct vertical top-to-bottom order [OK]

Common Mistakes:

• Assuming DFS preserves vertical order top-to-bottom
• Thinking Map keys must be strings
• Ignoring empty tree base case (which is handled here)

5. You want to perform vertical order traversal of a binary tree where nodes in the same column and row are sorted by value ascending. Which approach correctly achieves this?

hard

A. Use DFS and store nodes in a map by column only, then output map values.

B. Use BFS with a queue storing nodes with their row and column, then sort nodes by column, row, and value.

C. Traverse tree inorder and group nodes by column index only.

D. Use BFS but ignore row information and output nodes by column only.

Solution

1. Step 1: Understand sorting requirements

   Nodes must be sorted by column, then row (top to bottom), then by node value ascending if same row and column.

2. Step 2: Choose traversal and data structure

   BFS with queue tracks nodes level by level (row), and storing column and row allows sorting by these keys plus value.

3. Step 3: Why other options fail

   DFS does not guarantee row order. Inorder traversal does not track columns properly. Ignoring row info loses sorting order.

4. Final Answer:

   Use BFS with a queue storing nodes with their row and column, then sort nodes by column, row, and value. -> Option B

5. Quick Check:

   BFS + sort by col, row, value = correct order [OK]

Hint: Track row and column with BFS, then sort by col, row, value [OK]

Common Mistakes:

• Using DFS which loses row order
• Not sorting nodes by value when row and column are same
• Ignoring row info in BFS