Maximum Width of Binary Tree in DSA Javascript - Time & Space Complexity

Time Complexity: Maximum Width of Binary Tree

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the maximum width of a binary tree changes as the tree grows.

The question is: how does the work increase when the tree has more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function widthOfBinaryTree(root) {
  if (!root) return 0;
  let maxWidth = 0;
  let queue = [[root, 0]]; // node with index
  while (queue.length) {
    const length = queue.length;
    const start = queue[0][1];
    let end = start;
    for (let i = 0; i < length; i++) {
      const [node, index] = queue.shift();
      end = 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, end - start + 1);
  }
  return maxWidth;
}
    

This code finds the maximum width of a binary tree by level-order traversal, tracking node positions.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop processes each node once, and the inner for loop visits all nodes at the current level.
• How many times: Each node is visited exactly once, so total operations grow with the number of nodes.

How Execution Grows With Input

As the number of nodes (n) increases, the code visits each node once, so the work grows directly with n.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The operations increase linearly as the tree size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the maximum width grows in direct proportion to the number of nodes in the tree.

Common Mistake

[X] Wrong: "The time depends on the tree height only because we process level by level."

[OK] Correct: Even though we go level by level, every node is visited once, so total time depends on total nodes, not just height.

Interview Connect

Understanding how to analyze traversal algorithms like this helps you explain your approach clearly and shows you grasp how data size affects performance.

Self-Check

"What if we used a depth-first search instead of breadth-first search? How would the time complexity change?"

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