Maximum Width of Binary Tree in DSA Javascript

Mental Model

We want to find the widest level in a tree, counting all nodes between the leftmost and rightmost nodes including gaps.

Analogy: Imagine a theater seating where some seats are empty. The width is the distance from the first occupied seat on the left to the last occupied seat on the right, counting empty seats in between.

       1
      / \
     3   2
    /     \
   5       9
  /         \
 6           7

Levels:
Level 0: 1
Level 1: 3 -> 2
Level 2: 5 -> null -> null -> 9
Level 3: 6 -> null -> null -> null -> null -> null -> null -> 7

Dry Run Walkthrough

Input: Binary tree: 1 -> 3,2 -> 5,null,null,9 -> 6,null,null,null,null,null,null,7

Goal: Find the maximum width among all levels, counting nulls between nodes

Step 1: Start at root (1) with index 0

Level 0: [1(0)]

Why: We assign index 0 to root to track positions

Step 2: Move to level 1: nodes 3 and 2 with indices 1 and 2

Level 1: [3(1)] -> [2(2)]

Why: Left child index = 2*parent_index+1, right child = 2*parent_index+2

Step 3: Move to level 2: nodes 5 and 9 with indices 3 and 6 (nulls in between)

Level 2: [5(3)] -> null -> null -> [9(6)]

Why: Indices show gaps where null children exist

Step 4: Move to level 3: nodes 6 and 7 with indices 7 and 14 (many nulls between)

Level 3: [6(7)] -> null -> null -> null -> null -> null -> null -> [7(14)]

Why: Indices grow exponentially, capturing position gaps

Step 5: Calculate width per level as last_index - first_index + 1

Widths: Level 0=1, Level 1=2, Level 2=4, Level 3=8

Why: Width counts all positions including nulls between nodes

Step 6: Maximum width found is 8 at level 3

Result: Maximum width = 8

Why: Level 3 has the widest spread including gaps

Result:

Maximum width = 8

Annotated Code

DSA Javascript

class TreeNode {
  constructor(val = 0, left = null, right = null) {
    this.val = val;
    this.left = left;
    this.right = right;
  }
}

function widthOfBinaryTree(root) {
  if (!root) return 0;
  let maxWidth = 0;
  // Queue holds pairs: [node, index]
  const queue = [[root, 0]];

  while (queue.length > 0) {
    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 + 1]);
      if (node.right) queue.push([node.right, 2 * index + 2]);
    }
    // Width = last index - first index + 1
    const width = levelIndices[levelIndices.length - 1] - levelIndices[0] + 1;
    if (width > maxWidth) maxWidth = width;
  }
  return maxWidth;
}

// Driver code to build tree and test
const root = new TreeNode(1,
  new TreeNode(3,
    new TreeNode(5,
      new TreeNode(6), null
    ), null
  ),
  new TreeNode(2,
    null,
    new TreeNode(9,
      null,
      new TreeNode(7)
    )
  )
);

console.log(widthOfBinaryTree(root));

const queue = [[root, 0]];

Initialize queue with root node and index 0 to track positions

for (let i = 0; i < levelLength; i++) { const [node, index] = queue.shift();

Process each node in current level with its index

if (node.left) queue.push([node.left, 2 * index + 1]);

Assign left child index as 2*parent_index+1

if (node.right) queue.push([node.right, 2 * index + 2]);

Assign right child index as 2*parent_index+2

const width = levelIndices[levelIndices.length - 1] - levelIndices[0] + 1;

Calculate width of current level including gaps

if (width > maxWidth) maxWidth = width;

Update maximum width if current level is wider

Output

8

Complexity Analysis

Time: O(n) because we visit each node once in a breadth-first manner

Space: O(n) because the queue can hold up to all nodes in the widest level

vs Alternative: Compared to naive approaches that might store full tree structure or use recursion, this uses indexing to efficiently track positions and gaps without extra space for nulls

Edge Cases

Empty tree

Returns 0 as there are no nodes

DSA Javascript

if (!root) return 0;

Tree with only root node

Returns 1 as width is just the root

DSA Javascript

const queue = [[root, 0]];

Tree with single branch (all left or all right)

Width is always 1 since no siblings at any level

DSA Javascript

if (node.left) queue.push([node.left, 2 * index + 1]); if (node.right) queue.push([node.right, 2 * index + 2]);

When to Use This Pattern

When asked for the widest level in a tree including gaps, use level order traversal with indexing to track positions and calculate width efficiently.

Common Mistakes

Mistake: Not using indices to track positions, leading to incorrect width when nodes are missing in between
Fix: Assign indices to nodes as if the tree was a complete binary tree to count gaps correctly

Mistake: Calculating width as number of nodes in level instead of distance between leftmost and rightmost nodes
Fix: Calculate width as last_index - first_index + 1 using stored indices

Summary

Finds the maximum width of a binary tree by counting all nodes and gaps between the leftmost and rightmost nodes at each level.

Use when you need to measure the widest spread of nodes in a tree level including empty positions.

The key insight is to assign indices to nodes as if the tree was complete, so gaps are counted in width calculation.

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