Vertical Order Traversal of Binary Tree in DSA Javascript - Execution Trace

Concept Flow - Vertical Order Traversal of Binary Tree

Start at root node

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Assign horizontal distance = 0

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Use queue for BFS traversal

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Dequeue node

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Add node value to map at horizontal distance

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Enqueue left child with hd-1

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Enqueue right child with hd+1

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Repeat until queue empty

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Sort map keys

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Output values grouped by sorted keys

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Done

Traverse the tree level by level, track horizontal distances, group nodes by these distances, then output groups from left to right.

Execution Sample

DSA Javascript

function verticalOrder(root) {
  const map = new Map();
  const queue = [[root, 0]];
  while(queue.length) {
    const [node, hd] = queue.shift();
    if(!map.has(hd)) map.set(hd, []);
    map.get(hd).push(node.val);
    if(node.left) queue.push([node.left, hd - 1]);
    if(node.right) queue.push([node.right, hd + 1]);
  }
  return [...map.keys()].sort((a,b) => a-b).map(k => map.get(k));
}

This code performs vertical order traversal by BFS, grouping nodes by horizontal distance.

Execution Table

Step	Operation	Node Visited (val)	Horizontal Distance (hd)	Map State (hd: [values])	Queue State	Visual State
1	Start with root	1	0	0: [1]	[[2, -1], [3, 1]]	Root node 1 at hd=0
2	Visit node	2	-1	-1: [2], 0: [1]	[[3, 1], [4, -2], [5, 0]]	Left child 2 at hd=-1
3	Visit node	3	1	-1: [2], 0: [1], 1: [3]	[[4, -2], [5, 0], [6, 0], [7, 2]]	Right child 3 at hd=1
4	Visit node	4	-2	-2: [4], -1: [2], 0: [1], 1: [3]	[[5, 0], [6, 0], [7, 2]]	Left child 4 at hd=-2
5	Visit node	5	0	-2: [4], -1: [2], 0: [1,5], 1: [3]	[[6, 0], [7, 2]]	Node 5 at hd=0
6	Visit node	6	0	-2: [4], -1: [2], 0: [1,5,6], 1: [3]	[[7, 2]]	Node 6 at hd=0
7	Visit node	7	2	-2: [4], -1: [2], 0: [1,5,6], 1: [3], 2: [7]	[]	Right child 7 at hd=2
8	Queue empty	N/A	N/A	-2: [4], -1: [2], 0: [1,5,6], 1: [3], 2: [7]	[]	Traversal complete

💡 Queue is empty, all nodes visited.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	Final
queue	[[1,0]]	[[2,-1],[3,1]]	[[3,1],[4,-2],[5,0]]	[[4,-2],[5,0],[6,0],[7,2]]	[[5,0],[6,0],[7,2]]	[[6,0],[7,2]]	[[7,2]]	[]	[]
map	{}	{0:[1]}	{0:[1], -1:[2]}	{0:[1], -1:[2], 1:[3]}	{0:[1], -1:[2], 1:[3], -2:[4]}	{0:[1,5], -1:[2], 1:[3], -2:[4]}	{0:[1,5,6], -1:[2], 1:[3], -2:[4]}	{0:[1,5,6], -1:[2], 1:[3], -2:[4], 2:[7]}	{0:[1,5,6], -1:[2], 1:[3], -2:[4], 2:[7]}

Key Moments - 3 Insights

Why do we use a queue with nodes paired with horizontal distances?

The queue helps us visit nodes level by level (BFS). Pairing nodes with horizontal distances lets us group nodes vertically. See execution_table steps 1-7 where queue holds nodes with their hd.

Why do we add left child with hd-1 and right child with hd+1?

Left child is one step left, so hd decreases by 1; right child is one step right, so hd increases by 1. This is shown in execution_table steps 2-7 where hd changes accordingly.

How do we know when traversal is complete?

When the queue is empty, all nodes have been visited. Execution_table step 8 shows queue empty and traversal done.

Visual Quiz - 3 Questions

Look at the execution_table at Step 5, what is the map state?

A -2: [4], -1: [2], 0: [1,5], 1: [3]

B 0: [1,5], 1: [3], 2: [7]

C -1: [2], 0: [1], 1: [3]

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

💡 Hint

Check the 'Map State' column in execution_table row for Step 5.

At which step does the queue become empty?

A Step 6

B Step 7

C Step 8

D Step 5

💡 Hint

Look at the 'Queue State' column in execution_table; empty queue is shown at Step 8.

If we swapped left and right child enqueue order, how would the map values at hd=0 change?

A They would appear in reverse order [1,6,5]

B They would remain [1,5,6]

C Only the first value would change

D Map keys would change

💡 Hint

Check variable_tracker for map values at hd=0 after steps 5 and 6.

Concept Snapshot

Vertical Order Traversal:
- Use BFS with queue storing (node, horizontal distance)
- Left child hd = parent hd - 1
- Right child hd = parent hd + 1
- Group nodes by hd in map
- Sort keys and output grouped nodes
- Result shows nodes vertically from left to right

Full Transcript

Vertical Order Traversal visits nodes level by level, tracking their horizontal distance from the root. We use a queue to hold nodes paired with their horizontal distance. Starting at root with hd=0, we add nodes to a map grouped by hd. Left children get hd-1, right children hd+1. After visiting all nodes, we sort the map keys and output the grouped nodes. This shows the tree nodes vertically from left to right.

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