Maximum Width of Binary Tree in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Maximum Width Calculation

What is the output of the following code that calculates the maximum width of a binary tree?

DSA Javascript

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

function widthOfBinaryTree(root) {
  if (!root) return 0;
  let maxWidth = 0;
  let queue = [[root, 0]]; // node and index

  while (queue.length > 0) {
    const levelLength = queue.length;
    const levelIndices = queue.map(pair => pair[1]);
    const minIndex = levelIndices[0];
    const maxIndex = levelIndices[levelIndices.length - 1];
    maxWidth = Math.max(maxWidth, maxIndex - minIndex + 1);

    for (let i = 0; i < levelLength; i++) {
      const [node, index] = queue.shift();
      if (node.left) queue.push([node.left, 2 * (index - minIndex)]);
      if (node.right) queue.push([node.right, 2 * (index - minIndex) + 1]);
    }
  }
  return maxWidth;
}

// Tree structure:
//       1
//      / \
//     3   2
//    /     \
//   5       9
//  /         \
// 6           7

const root = new TreeNode(1,
  new TreeNode(3,
    new TreeNode(5,
      new TreeNode(6)
    )
  ),
  new TreeNode(2,
    null,
    new TreeNode(9,
      null,
      new TreeNode(7)
    )
  )
);

console.log(widthOfBinaryTree(root));

A 4

B 8

C 5

D 6

💡 Hint

Consider how the indices are assigned to nodes and how the width is calculated as the difference between max and min indices at each level.

The maximum width is calculated by tracking the indices of nodes at each level. The widest level here includes nodes at indices 0 to 7 (after normalization), so the width is 8.

❓ Predict Output

intermediate

Maximum Width for a Skewed Tree

What is the output of the code below for a skewed binary tree (all nodes only have a right child)?

DSA Javascript

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

function widthOfBinaryTree(root) {
  if (!root) return 0;
  let maxWidth = 0;
  let queue = [[root, 0]];

  while (queue.length > 0) {
    const levelLength = queue.length;
    const levelIndices = queue.map(pair => pair[1]);
    const minIndex = levelIndices[0];
    const maxIndex = levelIndices[levelIndices.length - 1];
    maxWidth = Math.max(maxWidth, maxIndex - minIndex + 1);

    for (let i = 0; i < levelLength; i++) {
      const [node, index] = queue.shift();
      if (node.left) queue.push([node.left, 2 * (index - minIndex)]);
      if (node.right) queue.push([node.right, 2 * (index - minIndex) + 1]);
    }
  }
  return maxWidth;
}

// Tree structure:
// 1 -> 2 -> 3 -> 4 (all right children)
const root = new TreeNode(1,
  null,
  new TreeNode(2,
    null,
    new TreeNode(3,
      null,
      new TreeNode(4)
    )
  )
);

console.log(widthOfBinaryTree(root));

A 3

B 2

C 4

D 1

💡 Hint

In a skewed tree, the width at each level is usually 1 because there is only one node per level.

Since each level has only one node, the width at every level is 1, so the maximum width is 1.

🔧 Debug

advanced

Identify the Error in Width Calculation

The following code attempts to calculate the maximum width of a binary tree but produces incorrect results. What is the main reason for the error?

DSA Javascript

function widthOfBinaryTree(root) {
  if (!root) return 0;
  let maxWidth = 0;
  let queue = [[root, 0]];

  while (queue.length > 0) {
    const levelLength = queue.length;
    const minIndex = queue[0][1];
    const maxIndex = queue[levelLength - 1][1];
    maxWidth = Math.max(maxWidth, maxIndex - minIndex + 1);

    for (let i = 0; i < levelLength; i++) {
      const [node, index] = queue.shift();
      if (node.left) queue.push([node.left, 2 * index]);
      if (node.right) queue.push([node.right, 2 * index + 1]);
    }
  }
  return maxWidth;
}

A The maxWidth calculation should use multiplication instead of subtraction.

B The queue is not initialized properly with root and index 1 instead of 0.

C Indices are not normalized by subtracting minIndex, causing overflow and incorrect width.

D The function does not handle null nodes in the queue, causing runtime errors.

💡 Hint

Check how indices are assigned to child nodes relative to the current level's minimum index.

Without normalizing indices by subtracting minIndex, indices can grow very large, causing incorrect width calculations due to integer overflow or gaps.

🧠 Conceptual

advanced

Why Use Index Normalization in Maximum Width Calculation?

Why is it important to normalize indices by subtracting the minimum index at each level when calculating the maximum width of a binary tree?

A To prevent integer overflow and keep indices small for accurate width calculation.

B To ensure the tree is balanced before calculating width.

C To sort the nodes at each level by their values.

D To convert the tree into a linked list for easier traversal.

💡 Hint

Think about what happens to indices as the tree depth increases.

Indices can become very large in deep trees, causing overflow or incorrect width calculations. Normalizing keeps indices relative and small.

🚀 Application

expert

Maximum Width of Binary Tree with Missing Nodes

Given the following binary tree, what is the maximum width calculated by the standard algorithm that uses index normalization? // Tree structure: // 1 // / \ // 2 3 // \ \ // 5 7 // / \ // 9 8

DSA Javascript

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

function widthOfBinaryTree(root) {
  if (!root) return 0;
  let maxWidth = 0;
  let queue = [[root, 0]];

  while (queue.length > 0) {
    const levelLength = queue.length;
    const minIndex = queue[0][1];
    const maxIndex = queue[levelLength - 1][1];
    maxWidth = Math.max(maxWidth, maxIndex - minIndex + 1);

    for (let i = 0; i < levelLength; i++) {
      const [node, index] = queue.shift();
      if (node.left) queue.push([node.left, 2 * (index - minIndex)]);
      if (node.right) queue.push([node.right, 2 * (index - minIndex) + 1]);
    }
  }
  return maxWidth;
}

const root = new TreeNode(1,
  new TreeNode(2,
    null,
    new TreeNode(5,
      new TreeNode(9)
    )
  ),
  new TreeNode(3,
    null,
    new TreeNode(7,
      null,
      new TreeNode(8)
    )
  )
);

console.log(widthOfBinaryTree(root));

A 6

B 5

C 8

D 7

💡 Hint

Consider the indices assigned to nodes at the third level and how gaps affect width.

At the third level, nodes 9 and 8 are at indices 0 and 5 respectively after normalization, so width is 6 (5 - 0 + 1). This is the maximum width.

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