Maximum Width of Binary Tree in DSA Javascript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the "maximum width" of a binary tree mean?

The maximum width of a binary tree is the maximum value of (maximum node index - minimum node index + 1) across all levels, where indices are assigned as in a complete binary tree.

intermediate

How can we keep track of node positions to calculate the maximum width of a binary tree?

We assign an index to each node as if the tree was a complete binary tree. The root is index 0, left child is 2 * parent_index + 1, right child is 2 * parent_index + 2. This helps calculate width by subtracting minimum from maximum indices at each level.

intermediate

Why do we subtract the minimum index from the maximum index at each level when finding the width?

The width of a level is calculated as (maximum index - minimum index + 1). Subtracting minimum from maximum and adding 1 accounts for all positions from leftmost to rightmost node, including gaps.

beginner

What data structure is commonly used to traverse the binary tree level by level for width calculation?

A queue is used for level order traversal (breadth-first search) to visit nodes level by level and track their indices.

intermediate

What is the time complexity of finding the maximum width of a binary tree using level order traversal?

The time complexity is O(n), where n is the number of nodes, because each node is visited once during the traversal.

What is the index of the right child if the parent node has index 3?

A 6

B 7

C 8

D 5

Right child index = 2 * parent_index + 2 = 2 * 3 + 2 = 8.

Which traversal method is best suited to calculate the maximum width of a binary tree?

A Preorder traversal

B Level order traversal

C Inorder traversal

D Postorder traversal

Level order traversal visits nodes level by level, which is needed to calculate width at each level.

If a level has nodes with indices 4, 5, and 7, what is the width of that level?

A 4

B 3

C 2

D 5

Width = max_index - min_index + 1 = 7 - 4 + 1 = 4.

What does the maximum width of a binary tree include?

A Only nodes present at the widest level

B Height of the tree

C Total number of nodes in the tree

D Nodes plus empty positions between them at the widest level

Maximum width counts nodes plus empty positions (gaps) between nodes at the widest level.

What is the initial index assigned to the root node when calculating maximum width?

A 1

B -1

C 0

D Depends on tree size

The root node is assigned index 0 to start the indexing system.

Explain how to calculate the maximum width of a binary tree using level order traversal and node indexing.

Think about numbering nodes like positions in a complete binary tree.

Describe why indexing nodes helps in finding the maximum width of a binary tree including gaps between nodes.

Imagine nodes placed on a number line with spaces.

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