Maximum Width of Binary Tree in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the queue with the root node and its index.

DSA Javascript

let queue = [[root, [1]]];

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A0

Bnull

C-1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting index at 1 instead of 0

Using null as index

Using negative index

The root node is assigned index 0 to start the level indexing from zero.

2fill in blank

medium

Complete the code to calculate the width of the current level.

DSA Javascript

let width = lastIndex - firstIndex + [1];

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A0

B-1

C1

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Not adding 1 to the difference

Subtracting instead of adding

Using zero which undercounts width

Width is calculated as last index minus first index plus one to include both ends.

3fill in blank

hard

Fix the error in the code that adds the left child to the queue with correct index.

DSA Javascript

if (node.left) queue.push([node.left, [1]]);

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Aindex * 2

Bindex * 2 + 1

Cindex + 1

Dindex - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using index * 2 + 1 for left child

Using index + 1 which is incorrect for tree indexing

Using negative indices

Left child's index is parent's index times 2 to maintain position in a full binary tree.

4fill in blank

hard

Fill both blanks to add the right child to the queue with correct index and update maxWidth.

DSA Javascript

if (node.right) queue.push([node.right, [1]]);
maxWidth = Math.max(maxWidth, [2]);

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Aindex * 2 + 1

Bwidth

ClastIndex - firstIndex + 1

Dindex * 2

Attempts:

3 left

💡 Hint

Common Mistakes

Using index * 2 for right child

Updating maxWidth with index instead of width

Not updating maxWidth at all

Right child's index is parent's index times 2 plus 1; maxWidth updates with current level width.

5fill in blank

hard

Fill all three blanks to complete the function that returns the maximum width of a binary tree.

DSA Javascript

function widthOfBinaryTree(root) {
  if (!root) return 0;
  let maxWidth = 0;
  let queue = [[root, [1]]];
  while (queue.length) {
    let levelLength = queue.length;
    let firstIndex = queue[0][1];
    let lastIndex = queue[queue.length - 1][1];
    maxWidth = Math.max(maxWidth, lastIndex - firstIndex + [2]);
    for (let i = 0; i < levelLength; i++) {
      let [node, index] = queue.shift();
      index -= firstIndex;
      if (node.left) queue.push([node.left, index * 2 + [3]]);
      if (node.right) queue.push([node.right, index * 2 + 1]);
    }
  }
  return maxWidth;
}

Drag options to blanks, or click blank then click option'

A0

B1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting index at 1

Not adding 1 to width

Using 1 instead of 0 for left child index offset

Start indexing at 0; add 1 to width calculation; left child index is index * 2 + 0.