medium Maximum Width of Binary Tree

Problem Statement

Given the root of a binary tree, find the maximum width of the tree.

The maximum width is the widest level in the tree.

The width of a level is the number of nodes between the leftmost and rightmost non-null nodes, where the null nodes between the end-nodes that would be present in a complete binary tree extending down to that level are also counted into the length calculation.

You can assume that the result will fit within a 32-bit signed integer.

Examples

Example 1

• Input: root = [1, 2, 3, 4, null, null, 5]

• Output: 4
• Justification: The maximum width is at the last level between nodes 4 and 5. It counts four positions: [4, null, null, 5].

Example 2

• Input: root = [1, 2, 3, 4, null, 5, 6, null, 7]

• Output: 4
• Justification: The maximum width is between nodes 4 and 6 at level 3, counting four positions: [4, null, 5, 6].

Example 3

• Input: root = [1, 2, null, 3, 4, null, null, 5]

• Output: 2
• Justification: The maximum width is at the third level, between nodes 3 and 4. It counts two positions: [3, 4].

Constraints:

• The number of nodes in the tree is in the range [1, 3000].
• -100 <= Node.val <= 100

Solution

To solve this problem, we will use a level order traversal approach, where we traverse the tree level by level from left to right. During traversal, we keep track of each node's position in a hypothetical "complete binary tree" to account for all gaps (nulls) between nodes at each level. We use a queue to store nodes along with their positions, which allows us to calculate the width of each level easily. At each level, we compare the positions of the leftmost and rightmost nodes to determine the width. The maximum width found across all levels is our answer.

This approach is effective because it leverages a breadth-first search (BFS) technique to explore each level fully before moving to the next, ensuring that we capture the correct width of every level. It is efficient in both time and space, given that it only requires storing nodes at one level at a time, making it well-suited for handling binary trees of reasonable size.

Step-by-Step Algorithm

• Check if the tree is empty:

  • If the root is null, return 0 as the width.

• Initialize a queue for level order traversal:

  • The queue will store pairs of nodes and their corresponding positions.

• Add the root node to the queue with position 0.

• Set a variable maxWidth to 0 to keep track of the maximum width found.

• While the queue is not empty:

  • Get the number of nodes at the current level (size).

  • Get the minimum position index at this level (minIndex) to normalize positions.

  • Initialize two variables first and last to store the positions of the first and last nodes at this level.

  • For each node at the current level:

    • Dequeue the node and its position from the queue.
    • Normalize the position by subtracting minIndex from the currentIndex to avoid overflow issues.
    • Update first with the index value if it is the first node at this level.
    • Update last with the index value if it is the last node at this level.
    • Check if the left child exists:
      • Add the left child to the queue with position 2 * current_position.
    • Check if the right child exists:
      • Add the right child to the queue with position 2 * current_position + 1.

  • Calculate the width of the current level as last - first + 1.

  • Update maxWidth to the maximum of its current value and the width of the current level.

• Return maxWidth after processing all levels.

Algorithm Walkthrough

Input: [1, 2, 3, 4, null, null, 5]

Given the binary tree:

    1
   / \
  2   3
 /     \
4       5

Initial Setup:

• Queue: [(1, 0)] (Node 1 at position 0)
• maxWidth: 0

Level 1

• Size: 1 (Only the root node 1)
• minIndex: 0 (Position of node 1)
• Initialize first and last to 0.

Process Node 1:

• Dequeue Node 1 with position 0.
• Update first to 0 (first node at this level).
• Update last to 0 (last node at this level).
• Enqueue left child (2) with position 0.
• Enqueue right child (3) with position 1.
• Calculate Width: last - first + 1 = 0 - 0 + 1 = 1
• Update maxWidth: max(0, 1) = 1.

Level 2

• Size: 2 (Nodes 2 and 3)
• minIndex: 0 (Position of node 2)
• Initialize first and last to 0.

Process Node 2:

• Dequeue Node 2 with position 0.
• Update first to 0 (first node at this level).
• Enqueue left child (4) with position 0 (no right child).

Process Node 3:

• Dequeue Node 3 with position 1.
• Update last to 1 (last node at this level).
• Enqueue right child (5) with position 2*index + 1 = 2*1 + 1 = 3 (no left child).
• Calculate Width: last - first + 1 = 1 - 0 + 1 = 2
• Update maxWidth: max(1, 2) = 2.

Level 3

• Size: 2 (Nodes 4 and 5)
• minIndex: 0 (Position of node 4)
• Initialize first and last to 0.

Process Node 4:

• Dequeue Node 4 with position 0.
• Update first to 0 (first node at this level).

Process Node 5:

• Dequeue Node 5 with position 3.
• Update last to 3 (last node at this level).
• Calculate Width: last - first + 1 = 3 - 0 + 1 = 4
• Update maxWidth: max(2, 4) = 4.

End

• The queue is empty; all levels are processed.
• Return maxWidth: 4.

Final Output: The maximum width of the binary tree is 4.

Code

import java.util.LinkedList;
import java.util.Queue;

// class TreeNode {
//     int val;
//     TreeNode left;
//     TreeNode right;

//     // Constructor to create a new tree node
//     TreeNode(int x) {
//         val = x;
//         left = null;
//         right = null;
//     }
// }

class Pair {

  TreeNode node;
  int index;

  // Constructor to create a new pair
  Pair(TreeNode node, int index) {
    this.node = node;
    this.index = index;
  }
}

class Solution {

  // Method to find the maximum width of the binary tree
  public int widthOfBinaryTree(TreeNode root) {
    if (root == null) return 0;

    Queue<Pair> queue = new LinkedList<>();
    queue.offer(new Pair(root, 0)); // Add the root node with position 0
    int maxWidth = 0; // Initialize maximum width

    // Level order traversal using queue
    while (!queue.isEmpty()) {
      int size = queue.size();
      int minIndex = queue.peek().index; // Get the minimum index at this level to normalize positions
      int first = 0, last = 0;

      for (int i = 0; i < size; i++) {
        Pair current = queue.poll();
        TreeNode node = current.node;
        int index = current.index - minIndex; // Normalize the index to prevent overflow

        // Record the first and last positions at the current level
        if (i == 0) first = index;
        if (i == size - 1) last = index;

        // Enqueue left child with the correct position if it exists
        if (node.left != null) {
          queue.offer(new Pair(node.left, 2 * index));
        }

        // Enqueue right child with the correct position if it exists
        if (node.right != null) {
          queue.offer(new Pair(node.right, 2 * index + 1));
        }
      }

      // Calculate the width of the current level and update maxWidth
      maxWidth = Math.max(maxWidth, last - first + 1);
    }

    return maxWidth;
  }

  public static void main(String[] args) {
    Solution sol = new Solution();

    // Create the first example tree
    TreeNode example1 = new TreeNode(1);
    example1.left = new TreeNode(2);
    example1.right = new TreeNode(3);
    example1.left.left = new TreeNode(4);
    example1.right.right = new TreeNode(5);

    // Create the second example tree
    TreeNode example2 = new TreeNode(1);
    example2.left = new TreeNode(2);
    example2.right = new TreeNode(3);
    example2.left.left = new TreeNode(4);
    example2.left.left.right = new TreeNode(7);
    example2.right.left = new TreeNode(5);
    example2.right.right = new TreeNode(6);

    // Create the third example tree
    TreeNode example3 = new TreeNode(1);
    example3.left = new TreeNode(2);
    example3.left.left = new TreeNode(3);
    example3.left.right = new TreeNode(4);
    example3.left.right.left = new TreeNode(5);

    // Test the widthOfBinaryTree method with the example trees
    System.out.println(sol.widthOfBinaryTree(example1)); // Output: 4
    System.out.println(sol.widthOfBinaryTree(example2)); // Output: 4
    System.out.println(sol.widthOfBinaryTree(example3)); // Output: 2
  }
}

Complexity Analys

Time Complexity

• The time complexity of this solution is , where N is the total number of nodes in the binary tree.
• This is because we perform a level-order traversal (BFS) of the tree, visiting each node exactly once to determine the maximum width.

Space Complexity

• The space complexity is , where W is the maximum width of the binary tree.
• The space complexity is determined by the maximum number of nodes that could be stored in the queue at any given level of the tree. In the worst case, this will be the width of the widest level of the binary tree.