Maximum Width of Binary Tree in DSA Javascript - Build from Scratch

Maximum Width of Binary Tree

📖 Scenario: You are working on a program that analyzes the structure of a binary tree. The goal is to find the maximum width of the tree, which means the largest number of nodes present at any single level.Imagine the tree as a family tree where each generation is a level. You want to find the generation with the most family members.

🎯 Goal: Build a JavaScript program that calculates the maximum width of a given binary tree. You will create the tree nodes, set up a queue for level order traversal, calculate the width at each level, and finally print the maximum width.

📋 What You'll Learn

Create a binary tree using nodes with val, left, and right properties

Use a queue to perform level order traversal

Calculate the width of each level by counting nodes

Find and print the maximum width among all levels

💡 Why This Matters

🌍 Real World

Finding the maximum width of a binary tree helps in understanding the structure and balance of hierarchical data, such as organizational charts or file systems.

💼 Career

This concept is useful for software engineers working with tree data structures, optimizing algorithms, and solving problems related to data organization and traversal.

1

Create the Binary Tree Nodes

Create a class called TreeNode with a constructor that takes val and sets this.val, this.left, and this.right to null. Then create a binary tree with root node value 1, root's left child value 3, root's right child value 2, left child's left child value 5, left child's right child value 3, and right child's left child value 4.

DSA Javascript

# Your code here

Hint

Define the TreeNode class first, then create the nodes and link them as described.

2

Set Up a Queue for Level Order Traversal

Create a variable called queue and initialize it as an array containing the root node. Also create a variable called maxWidth and set it to 0 to keep track of the maximum width found.

DSA Javascript

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

const root = new TreeNode(1);
root.left = new TreeNode(3);
root.right = new TreeNode(2);
root.left.left = new TreeNode(5);
root.left.right = new TreeNode(3);
root.right.left = new TreeNode(4);

// Initialize queue and maxWidth
# Your code here

Hint

Use an array to hold nodes for level order traversal and a variable to track the maximum width.

3

Calculate Maximum Width Using Level Order Traversal

Use a while loop that runs as long as queue.length is greater than 0. Inside the loop, create a variable levelLength set to queue.length. Update maxWidth to be the maximum of maxWidth and levelLength. Then use a for loop from 0 to levelLength - 1 to remove the first node from queue and add its left and right children to queue if they exist.

DSA Javascript

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

const root = new TreeNode(1);
root.left = new TreeNode(3);
root.right = new TreeNode(2);
root.left.left = new TreeNode(5);
root.left.right = new TreeNode(3);
root.right.left = new TreeNode(4);

const queue = [root];
let maxWidth = 0;

// Calculate maxWidth using level order traversal
# Your code here

Hint

Use a queue to process each level, update maxWidth, and add children nodes for the next level.

4

Print the Maximum Width

Write a console.log statement to print the value of maxWidth.

DSA Javascript

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

const root = new TreeNode(1);
root.left = new TreeNode(3);
root.right = new TreeNode(2);
root.left.left = new TreeNode(5);
root.left.right = new TreeNode(3);
root.right.left = new TreeNode(4);

const queue = [root];
let maxWidth = 0;

while (queue.length > 0) {
    const levelLength = queue.length;
    maxWidth = Math.max(maxWidth, levelLength);
    for (let i = 0; i < levelLength; i++) {
        const node = queue.shift();
        if (node.left) queue.push(node.left);
        if (node.right) queue.push(node.right);
    }
}

// Print the maximum width
# Your code here

Hint

Use console.log(maxWidth) to display the final answer.

Practice

1. What does the "maximum width" of a binary tree represent?

easy

A. The height of the tree from root to the deepest leaf

B. The total number of nodes in the tree

C. The largest number of nodes present at any single level of the tree

D. The sum of values of all nodes at the widest level

Solution

1. Step 1: Understand the term 'width' in a binary tree

   The width refers to how many nodes exist at the same depth or level in the tree.

2. Step 2: Identify what 'maximum width' means

   It is the largest count of nodes found at any one level across the entire tree.

3. Final Answer:

   The largest number of nodes present at any single level of the tree -> Option C

4. Quick Check:

   Maximum width = max nodes at any level [OK]

Hint: Width means count of nodes at one level, not height or sum [OK]

Common Mistakes:

• Confusing width with height of the tree
• Thinking width is total nodes in the tree
• Mixing width with sum of node values

2. Which of the following JavaScript code snippets correctly initializes a queue to help calculate the maximum width of a binary tree?

easy

A. const queue = {};

B. const queue = [];

C. const queue = new Set();

D. const queue = null;

Solution

1. Step 1: Identify the correct data structure for BFS traversal

   To calculate maximum width, we use level order traversal which requires a queue (FIFO).

2. Step 2: Check JavaScript syntax for queue initialization

   An array can be used as a queue with push and shift methods, so const queue = [] is correct.

3. Final Answer:

   const queue = []; -> Option B

4. Quick Check:

   Queue for BFS = array [] [OK]

Hint: Use array [] as queue for BFS in JavaScript [OK]

Common Mistakes:

• Using Set which is unordered
• Using object {} which is not a queue
• Assigning null instead of a data structure

3. Consider the following JavaScript code to find the maximum width of a binary tree. What will be the output?

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

const root = new TreeNode(1,
  new TreeNode(2, new TreeNode(4), new TreeNode(5)),
  new TreeNode(3, null, new TreeNode(7))
);

function widthOfBinaryTree(root) {
  if (!root) return 0;
  let maxWidth = 0;
  const queue = [[root, 0]]; // node with index
  while (queue.length) {
    const levelLength = queue.length;
    const levelIndices = [];
    for (let i = 0; i < levelLength; i++) {
      const [node, index] = queue.shift();
      levelIndices.push(index);
      if (node.left) queue.push([node.left, 2 * index]);
      if (node.right) queue.push([node.right, 2 * index + 1]);
    }
    maxWidth = Math.max(maxWidth, levelIndices[levelIndices.length - 1] - levelIndices[0] + 1);
  }
  return maxWidth;
}

console.log(widthOfBinaryTree(root));

medium

A. 4

B. 3

C. 2

D. 5

Solution

1. Step 1: Analyze the tree structure and node indices

   Level 0: Node 1 at index 0 (width 1) Level 1: Nodes 2 (index 0), 3 (index 1) width = 2 Level 2: Nodes 4 (0), 5 (1), null, 7 (3) width = 3 - 0 + 1 = 4

2. Step 2: Calculate maximum width from indices

   The widest level is level 2 with width 4, so maxWidth = 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Max width = 4 nodes at level 2 [OK]

Hint: Track node indices to find width = max index - min index + 1 [OK]

Common Mistakes:

• Ignoring null gaps in indexing
• Counting nodes instead of width using indices
• Confusing index calculation for children

4. The following code attempts to compute the maximum width of a binary tree but has a bug. What is the bug?

function maxWidth(root) {
  if (!root) return 0;
  let maxWidth = 0;
  const queue = [root];
  while (queue.length) {
    const levelLength = queue.length;
    maxWidth = Math.max(maxWidth, levelLength);
    for (let i = 0; i < levelLength; i++) {
      const node = queue.shift();
      if (node.left) queue.push(node.left);
      if (node.right) queue.push(node.right);
    }
  }
  return maxWidth;
}

medium

A. It pushes nodes into queue before checking if queue is empty

B. It uses shift() which is not allowed in JavaScript arrays

C. It does not initialize maxWidth to zero

D. It does not consider null nodes between children, so width may be underestimated

Solution

1. Step 1: Understand how width is calculated here

   This code counts nodes at each level but ignores gaps caused by missing nodes (nulls).

2. Step 2: Identify why ignoring nulls causes wrong width

   Width should count positions including nulls between nodes, but this code only counts existing nodes, underestimating width.

3. Final Answer:

   It does not consider null nodes between children, so width may be underestimated -> Option D

4. Quick Check:

   Missing nulls causes wrong width [OK]

Hint: Include nulls in queue to track gaps for correct width [OK]

Common Mistakes:

• Thinking shift() is invalid in JS arrays
• Ignoring null gaps in width calculation
• Misunderstanding queue initialization

5. You want to find the maximum width of a binary tree but the tree can be very deep and sparse. Which approach below best handles this efficiently in JavaScript?

hard

A. Use level order traversal with node indices to track positions and calculate width

B. Use depth-first traversal and count nodes at each depth without indices

C. Use recursion to count total nodes and divide by height

D. Use a stack to store nodes and calculate width by node values

Solution

1. Step 1: Consider the problem of deep and sparse trees

   Deep sparse trees have missing nodes causing gaps; simple node counts per level won't work.

2. Step 2: Choose method that tracks node positions

   Level order traversal with indices tracks exact positions, including gaps, allowing correct width calculation.

3. Step 3: Evaluate other options

   DFS without indices can't track gaps; recursion counting nodes/height is inaccurate; stack with node values doesn't help width.

4. Final Answer:

   Use level order traversal with node indices to track positions and calculate width -> Option A

5. Quick Check:

   Index tracking + BFS = correct max width [OK]

Hint: Track indices in BFS to handle sparse trees efficiently [OK]

Common Mistakes:

• Using DFS without position tracking
• Assuming total nodes/height gives width
• Using node values instead of positions