Maximum Width of Binary Tree in DSA Javascript - Execution Trace

Concept Flow - Maximum Width of Binary Tree

Start at root node

↓

Initialize queue with root and index 0

↓

While queue not empty

↓

For each level: get first and last node indices

↓

Calculate width = last_index - first_index + 1

↓

Update max width if current width is larger

↓

Add children to queue with updated indices

↓

Repeat until all levels processed

↓

Return max width found

We start from the root and use a queue to traverse level by level, tracking node positions to find the maximum width.

Execution Sample

DSA Javascript

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

This code finds the maximum width of a binary tree by tracking node indices level-wise.

Execution Table

Step	Operation	Nodes in Queue	Pointer Changes	Visual State
1	Initialize queue with root (index 0)	[[1,0]]	queue = [[1,0]]	Level 0: 1(0)
2	Process level 0: pop node 1 at index 0	[]	queue.shift() -> [1,0]	Level 0 processed
3	Add children of node 1: left=2 at index 1, right=3 at index 2	[[2,1],[3,2]]	queue.push([2,1]), queue.push([3,2])	Level 1: 2(1) -> 3(2)
4	Calculate width level 0: lastIndex 0 - firstIndex 0 + 1 = 1	[[2,1],[3,2]]	maxWidth = 1	Max width updated to 1
5	Process level 1: pop node 2 at index 1	[[3,2]]	queue.shift() -> [2,1]	Processing node 2
6	Add children of node 2: left=4 at index 3, right=5 at index 4	[[3,2],[4,3],[5,4]]	queue.push([4,3]), queue.push([5,4])	Level 2: 4(3) -> 5(4)
7	Process level 1: pop node 3 at index 2	[[4,3],[5,4]]	queue.shift() -> [3,2]	Processing node 3
8	Add children of node 3: none	[[4,3],[5,4]]	No children added	Level 2 unchanged
9	Calculate width level 1: lastIndex 2 - firstIndex 1 + 1 = 2	[[4,3],[5,4]]	maxWidth = 2	Max width updated to 2
10	Process level 2: pop node 4 at index 3	[[5,4]]	queue.shift() -> [4,3]	Processing node 4
11	Add children of node 4: none	[[5,4]]	No children added	Level 3 unchanged
12	Process level 2: pop node 5 at index 4	[]	queue.shift() -> [5,4]	Processing node 5
13	Add children of node 5: none	[]	No children added	Level 3 empty
14	Calculate width level 2: lastIndex 4 - firstIndex 3 + 1 = 2	[]	maxWidth = 2	Max width remains 2
15	Queue empty, end loop	[]	Return maxWidth = 2	Final max width = 2

💡 Queue is empty, all levels processed, maximum width found is 2

Variable Tracker

Variable	Start	After Step 3	After Step 6	After Step 9	After Step 14	Final
queue	[[1,0]]	[[2,1],[3,2]]	[[3,2],[4,3],[5,4]]	[[4,3],[5,4]]	[]	[]
maxWidth	0	1	1	2	2	2
firstIndex	0	1	1	1	-	-
lastIndex	0	2	4	4	-	-

Key Moments - 3 Insights

Why do we assign indices to nodes in the queue?

Indices help us calculate the width by showing the position of nodes in the level. See steps 3, 6, and 9 where indices are used to find width.

Why do we subtract firstIndex from lastIndex and add 1 to get width?

Because width counts all positions between first and last nodes inclusive. This is shown in steps 4, 9, and 14 in the execution table.

Why do we multiply index by 2 and add 1 or 2 for children?

This simulates the position of left and right children in a full binary tree, keeping track of gaps. See steps 3 and 6 for how children indices are calculated.

Visual Quiz - 3 Questions

Look at the execution table at step 9, what is the maxWidth value?

A 1

B 2

C 3

D 4

💡 Hint

Check the 'maxWidth' column in the execution table row for step 9.

At which step does the queue become empty, ending the loop?

A Step 14

B Step 13

C Step 15

D Step 10

💡 Hint

Look at the 'Queue' column in the execution table to find when it is empty.

If the root had no children, what would be the maximum width?

A 1

B 2

C 0

D Undefined

💡 Hint

Refer to the initial queue state and width calculation at step 4.

Concept Snapshot

Maximum Width of Binary Tree:
- Use level order traversal with a queue
- Assign indices to nodes to track positions
- Width = last index - first index + 1 per level
- Update max width after each level
- Return max width after processing all levels

Full Transcript

To find the maximum width of a binary tree, we start at the root and use a queue to traverse level by level. Each node is paired with an index representing its position if the tree was complete. For each level, we calculate the width by subtracting the first node's index from the last node's index and adding one. We update the maximum width if the current level's width is larger. We add children to the queue with calculated indices and repeat until all levels are processed. Finally, we return the maximum width found.

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