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

Time Complexity: Maximum Width of Binary Tree

O(n)

Understanding Time Complexity

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

How does the number of nodes affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int widthOfBinaryTree(TreeNode* root) {
    if (!root) return 0;
    unsigned long long maxWidth = 0;
    queue> q;
    q.push({root, 0});
    while (!q.empty()) {
        int size = q.size();
        unsigned long long start = q.front().second;
        unsigned long long end = q.back().second;
        maxWidth = max(maxWidth, end - start + 1);
        for (int i = 0; i < size; i++) {
            auto node = q.front().first;
            unsigned long long idx = q.front().second - start;
            q.pop();
            if (node->left) q.push({node->left, 2 * idx + 1});
            if (node->right) q.push({node->right, 2 * idx + 2});
        }
    }
    return (int)maxWidth;
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop runs once per level of the tree, and inside it, a for loop processes all nodes at that level.
• How many times: Each node is visited exactly once during the traversal.

How Execution Grows With Input

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

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

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

Final Time Complexity

Time Complexity: O(n)

This means the time to find the maximum width grows directly with the number of nodes in the tree.

Common Mistake

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

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

Interview Connect

Understanding how to analyze level-order traversal helps you explain many tree problems clearly and confidently in interviews.

Self-Check

"What if we used a depth-first search instead of a 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 term 'width' in a tree

   The width of a binary tree at a level is the count of nodes present at that level.

2. Step 2: Define 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 count [OK]

Common Mistakes:

• Confusing width with tree height
• Counting total nodes instead of per level
• Summing node values instead of counting nodes

2. Which of the following C++ code snippets correctly declares a queue to store pairs of TreeNode pointers and their indices for width calculation?

easy

A. queue> q;

B. std::queue q;

C. std::queue> q;

D. std::queue> q;

Solution

1. Step 1: Understand queue declaration with pairs

   To store pairs, use std::pair with template parameters inside std::queue<>.

2. Step 2: Check syntax correctness

   std::queue> q; correctly uses std::queue> q; which is valid C++ syntax.

3. Final Answer:

   std::queue> q; -> Option D

4. Quick Check:

   Queue of pairs syntax = std::queue> [OK]

Hint: Use std::pair inside std::queue with angle brackets [OK]

Common Mistakes:

• Missing std::pair<> inside queue
• Incorrect template syntax
• Omitting std:: prefix

3. Given the following binary tree and code snippet, what is the maximum width printed?

struct TreeNode {
  int val;
  TreeNode* left;
  TreeNode* right;
  TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

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

int maxWidth(TreeNode* root) {
  if (!root) return 0;
  std::queue> q;
  q.push({root, 0});
  int maxWidth = 0;
  while (!q.empty()) {
    int size = q.size();
    int start = q.front().second;
    int end = start;
    for (int i = 0; i < size; i++) {
      auto [node, idx] = q.front(); q.pop();
      end = idx;
      if (node->left) q.push({node->left, 2 * idx + 1});
      if (node->right) q.push({node->right, 2 * idx + 2});
    }
    maxWidth = std::max(maxWidth, end - start + 1);
  }
  return maxWidth;
}

// Call: maxWidth(root);

medium

A. 2

B. 3

C. 4

D. 5

Solution

1. Step 1: Analyze tree levels and node indices

   Level 0: Node 1 at index 0 (width = 1) Level 1: Nodes 2 (1), 3 (2) width = 2 Level 2: Nodes 4 (3), 5 (6) width = 6 - 3 + 1 = 4

2. Step 2: Calculate maximum width using code logic

   Code calculates width as end - start + 1 per level. Level 2 width is 4, which is maximum.

3. Final Answer:

   4 -> Option C

4. Quick Check:

   Max width = 4 nodes at level 2 [OK]

Hint: Track indices per level, width = end - start + 1 [OK]

Common Mistakes:

• Counting nodes without index gaps
• Ignoring index-based width calculation
• Assuming width equals node count only

4. Identify the bug in the following code snippet that attempts to compute the maximum width of a binary tree:

int widthOfBinaryTree(TreeNode* root) {
  std::queue> q;
  q.push({root, 0});
  int maxWidth = 0;
  while (!q.empty()) {
    int size = q.size();
    int start = q.front().second;
    int end = q.back().second;
    maxWidth = std::max(maxWidth, end - start + 1);
    for (int i = 0; i < size; i++) {
      auto [node, idx] = q.front(); q.pop();
      if (node->left) q.push({node->left, 2 * idx + 1});
      if (node->right) q.push({node->right, 2 * idx + 2});
    }
  }
  return maxWidth;
}

medium

A. Not checking if root is nullptr before processing

B. Using q.back().second before popping nodes causes incorrect width calculation

C. Incorrect indexing formula for child nodes

D. Missing return statement at the end

Solution

1. Step 1: Analyze the width calculation timing

   The code calculates width using q.back().second before popping nodes in the current level.

2. Step 2: Identify the bug

   Since q.back() includes nodes from the next level (pushed during previous iteration), width calculation is incorrect.

3. Final Answer:

   Using q.back().second before popping nodes causes incorrect width calculation -> Option B

4. Quick Check:

   Calculate width after popping all nodes at current level [OK]

Hint: Calculate width after processing current level nodes [OK]

Common Mistakes:

• Forgetting root null check
• Misplacing width calculation inside loop
• Wrong child index formulas

5. You want to find the maximum width of a binary tree that may have missing nodes between left and right children at some levels. Which approach correctly handles gaps caused by missing nodes?

hard

A. Use level order traversal with node indices to track positions and calculate width as end - start + 1

B. Count only the number of nodes at each level ignoring their positions

C. Calculate width as the difference between maximum and minimum node values at each level

D. Use depth-first traversal and count nodes recursively without indices

Solution

1. Step 1: Understand the problem with missing nodes

   Missing nodes create gaps, so counting nodes alone misses true width.

2. Step 2: Use indices to track node positions

   Assign indices to nodes as if in a full binary tree to capture gaps, then width = end - start + 1.

3. Final Answer:

   Use level order traversal with node indices to track positions and calculate width as end - start + 1 -> Option A

4. Quick Check:

   Index-based width handles gaps correctly [OK]

Hint: Track node indices to include gaps in width calculation [OK]

Common Mistakes:

• Ignoring gaps by counting nodes only
• Using node values instead of positions
• Using DFS without position tracking