Maximum Width of Binary Tree in DSA Java - 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.

Specifically, how does the number of nodes affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public int widthOfBinaryTree(TreeNode root) {
    if (root == null) return 0;
    Queue<Pair<TreeNode, Integer>> queue = new LinkedList<>();
    queue.offer(new Pair<>(root, 0));
    int maxWidth = 0;
    while (!queue.isEmpty()) {
        int size = queue.size();
        int minIndex = queue.peek().getValue();
        int first = 0, last = 0;
        for (int i = 0; i < size; i++) {
            Pair<TreeNode, Integer> p = queue.poll();
            TreeNode node = p.getKey();
            int index = p.getValue() - minIndex;
            if (i == 0) first = index;
            if (i == size - 1) last = index;
            if (node.left != null) queue.offer(new Pair<>(node.left, 2 * index + 1));
            if (node.right != null) queue.offer(new Pair<>(node.right, 2 * index + 2));
        }
        maxWidth = Math.max(maxWidth, last - first + 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 that visits each level of the tree and the inner for loop that processes each node at that level.
• How many times: Each node in the tree is visited exactly once during the traversal.

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 to nodes
100	About 100 visits to nodes
1000	About 1000 visits to nodes

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 complexity is O(width) because we only process nodes level by level."

[OK] Correct: Even though we process nodes level by level, every node in the tree is visited once, so the total work depends on the total number of nodes, not just the width.

Interview Connect

Understanding how tree traversals scale helps you explain your approach clearly and shows you can reason about efficiency, a key skill in coding interviews.

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 largest number of nodes present at any single level of the tree

B. The height of the tree from root to the deepest leaf

C. The total number of nodes in the tree

D. The sum of values of all nodes at the widest level

Solution

1. Step 1: Understand the definition of width in a binary tree

   The width is the count of nodes at a particular level in the tree.

2. Step 2: Identify maximum width

   The maximum width is the largest such count among all levels.

3. Final Answer:

   The largest number of nodes present at any single level of the tree -> Option A

4. Quick Check:

   Maximum width = max nodes at any level [OK]

Hint: Width means count of nodes at one level, find max [OK]

Common Mistakes:

• Confusing width with tree height
• Thinking width is sum of node values
• Counting total nodes instead of per level

2. Which Java data structure is most suitable to track nodes level by level when finding maximum width of a binary tree?

easy

A. Stack

B. HashMap

C. Queue

D. ArrayList

Solution

1. Step 1: Identify traversal method for level order

   Level order traversal uses a queue to process nodes level by level.

2. Step 2: Choose data structure

   A queue allows nodes to be added and removed in FIFO order, perfect for level traversal.

3. Final Answer:

   Queue -> Option C

4. Quick Check:

   Level order traversal uses queue [OK]

Hint: Use queue for level order traversal to measure width [OK]

Common Mistakes:

• Using stack which is LIFO, not suitable for level order
• Choosing HashMap which doesn't preserve order
• Using ArrayList without queue behavior

3. What is the output of the following Java code snippet that calculates maximum width of a binary tree?

class TreeNode {
  int val;
  TreeNode left, right;
  TreeNode(int v) { val = v; }
}

// Tree structure:
//       1
//      / \
//     2   3
//    /     \
//   4       5

TreeNode root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);
root.left.left = new TreeNode(4);
root.right.right = new TreeNode(5);

// Maximum width calculation
int maxWidth = 0;
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
  int size = queue.size();
  maxWidth = Math.max(maxWidth, size);
  for (int i = 0; i < size; i++) {
    TreeNode node = queue.poll();
    if (node.left != null) queue.add(node.left);
    if (node.right != null) queue.add(node.right);
  }
}
System.out.println(maxWidth);

medium

A. 1

B. 2

C. 3

D. 4

Solution

1. Step 1: Analyze tree levels and node counts

   Level 1 has 1 node (1), level 2 has 2 nodes (2,3), level 3 has 2 nodes (4,5).

2. Step 2: Determine maximum width

   The maximum number of nodes at any level is 2.

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Max width = 2 nodes at level 2 [OK]

Hint: Count nodes per level, max is answer [OK]

Common Mistakes:

• Counting total nodes instead of per level
• Assuming level 3 has 3 nodes incorrectly
• Ignoring null children in width calculation

4. Identify the bug in this Java code snippet that attempts to find the maximum width of a binary tree:

Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
int maxWidth = 0;
while (!queue.isEmpty()) {
  int size = queue.size();
  maxWidth = Math.max(maxWidth, size);
  for (int i = 0; i <= size; i++) {
    TreeNode node = queue.poll();
    if (node.left != null) queue.add(node.left);
    if (node.right != null) queue.add(node.right);
  }
}

medium

A. The loop condition should be i < size, not i <= size

B. The queue should be a Stack instead

C. maxWidth should be updated after the for loop

D. The root node is never added to the queue

Solution

1. Step 1: Check the for loop iteration count

   The loop runs from i = 0 to i <= size, which means it runs size+1 times, causing an extra poll.

2. Step 2: Correct the loop condition

   It should run exactly size times, so condition must be i < size.

3. Final Answer:

   The loop condition should be i < size, not i <= size -> Option A

4. Quick Check:

   Loop must run size times exactly [OK]

Hint: Use i < size for loops over queue size [OK]

Common Mistakes:

• Using <= instead of < in for loop
• Replacing queue with stack incorrectly
• Updating maxWidth at wrong place

5. Given a binary tree where some nodes are missing children, how can you modify the maximum width calculation to include null positions between nodes, so the width counts the distance between the leftmost and rightmost non-null nodes at each level? Choose the best approach.

hard

A. Calculate width by summing values of nodes at each level

B. Count only non-null nodes at each level ignoring positions

C. Use a stack to store nodes and count them

D. Assign an index to each node as if the tree was a complete binary tree, track min and max indices per level, width = max - min + 1

Solution

1. Step 1: Understand the problem with missing nodes

   Simple counting misses gaps caused by null children, so width is underestimated.

2. Step 2: Use indexing to simulate complete tree positions

   Assign indices to nodes as in a complete binary tree (root=0, left=2*index, right=2*index+1).

3. Step 3: Calculate width per level using min and max indices

   Width = max index - min index + 1, capturing gaps between nodes.

4. Final Answer:

   Assign an index to each node as if the tree was a complete binary tree, track min and max indices per level, width = max - min + 1 -> Option D

5. Quick Check:

   Indexing nodes captures gaps for accurate width [OK]

Hint: Use node indices to count gaps for true width [OK]

Common Mistakes:

• Ignoring null gaps between nodes
• Counting only non-null nodes without positions
• Using stack instead of queue for level traversal