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Maximum width of a Binary Tree with null values | Set 2

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  • Last Updated : 07 Apr, 2022

Pre-requisite: Maximum width of a Binary Tree with null value | Set 1

Given a Binary Tree consisting of N nodes, the task is to find the maximum width of the given tree without using recursion, where the maximum width is defined as the maximum of all the widths at each level of the given Tree.

Note: The width of a tree for any level is defined as the number of nodes between the two extreme nodes of that level including the NULL node in between.

Examples:

Input:
                       1        
                    /     \                                  
                  2       3         
              /     \       \                       
            4      5       8  
        /     \                                          
     6        7                     
Output: 4
Explanation:

                       1                  // width = 1
                    /     \                                  
                  2       3              // width = 2 – 1 + 1=  2
              /     \       \                       
            4      5       8            // width = 6 – 3 + 1 = 4
        /     \                                          
     6        7                          // width = 8 – 7 + 1 = 2

So, the answer is 4

Input:
                   1
                 /
              2 
            /
          3    
Output: 1

 

Approach: In this approach, the main idea is to use level order traversal and to give an id to all nodes according to their parent which will help to find the width of a particular level. Ids are distributed in this particular order:

                                     [node, id: i]
                                     /              \
  [left child, id: (i * 2 +1)]            [right child, id: (i * 2 + 2)]                

Now the width of each level can be calculated using the formula

Width of level i = (id of first node at level i) – (id of last node at level i) +1

Illustration: For example, use the first example provided here.

                       {1,0}                     // width = 1
                      /     \                                  
               {2,1}       {3,2}               // width = 2 – 1 + 1=  2
              /     \             \                       
      {4,3}      {5,4}         {8,6}         // width = 6 – 3 + 1 = 4
     /     \                                          
 {6,7}    {7,8}                               // width = 8 – 7 + 1 = 2

So, the answer is 4

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Tree Node structure
struct Node {
    int data;
    Node *left, *right;
 
    // Constructor
    Node(int item)
    {
        data = item;
        left = right = NULL;
    }
};
 
// Function to find the width
// of the given tree
int getMaxWidth(Node* root)
{
    if (root == NULL) {
        return 0;
    }
    queue<pair<Node*, int> > q;
    q.push({ root, 0 });
    int maxWidth = 1;
 
    while (!q.empty()) {
        int size = q.size();
        int first, last;
 
        for (int i = 0; i < size; i++) {
            Node* temp = q.front().first;
            int id = q.front().second;
            q.pop();
 
            // First Node
            if (i == 0) {
                first = id;
            }
 
            // Last Node
            if (i == size - 1) {
                last = id;
            };
 
            if (temp->left)
                q.push({ temp->left,
                        id * 2 + 1 });
            if (temp->right)
                q.push({ temp->right,
                        id * 2 + 2 });
        }
 
        maxWidth = max(maxWidth,
                       last - first + 1);
    }
    return maxWidth;
}
 
// Driver Code
int main()
{
    struct Node* root = new Node(1);
    root->left = new Node(2);
    root->right = new Node(3);
    root->left->left = new Node(4);
    root->left->right = new Node(5);
    root->right->right = new Node(8);
    root->right->right->left = new Node(6);
    root->right->right->right = new Node(7);
 
    // Function Call
    cout << getMaxWidth(root);
    return 0;
}


Java




// Java program for the above approach
 
import java.util.LinkedList;
import java.util.Queue;
 
class GFG {
 
    // Tree Node structure
    public static class Node {
        int data;
        Node left;
        Node right;
 
        // Constructor
        public Node(int item) {
            this.data = item;
            this.left = this.right = null;
        }
    };
 
    public static class Pair {
        Node first;
        int second;
 
        public Pair(Node n, int i) {
            this.first = n;
            this.second = i;
        }
    }
 
    // Function to find the width
    // of the given tree
    static int getMaxWidth(Node root) {
        if (root == null) {
            return 0;
        }
        Queue<Pair> q = new LinkedList<Pair>();
        q.add(new Pair(root, 0));
        int maxWidth = 1;
 
        while (!q.isEmpty()) {
            int size = q.size();
            int first = 0, last = 0;
 
            for (int i = 0; i < size; i++) {
                Node temp = q.peek().first;
                int id = q.peek().second;
                q.remove();
 
                // First Node
                if (i == 0) {
                    first = id;
                }
 
                // Last Node
                if (i == size - 1) {
                    last = id;
                }
                ;
 
                if (temp.left != null)
                    q.add(new Pair(temp.left, id * 2 + 1));
                if (temp.right != null)
                    q.add(new Pair(temp.right, id * 2 + 2));
            }
 
            maxWidth = Math.max(maxWidth, last - first + 1);
        }
        return maxWidth;
    }
 
    // Driver Code
    public static void main(String args[]) {
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);
        root.right.right = new Node(8);
        root.right.right.left = new Node(6);
        root.right.right.right = new Node(7);
 
        // Function Call
        System.out.println(getMaxWidth(root));
    }
}
 
// This code is contributed by saurabh_jaiswal.


Python3




# Python program for the above approach
 
# Tree Node structure
class Node:
 
    # Constructor
    def __init__(self, item):
        self.data = item
        self.left = self.right = None
 
# Function to find the width
# of the given tree
def getMaxWidth(root):
    if (root == None):
        return 0
 
    q = []
    q.append([root, 0])
    maxWidth = 1
 
    while (len(q)):
        size = len(q)
        first = None
        last = None
 
        for i in range(size):
            temp = q[0][0]
            id = q[0][1]
            q.pop(0)
 
            # First Node
            if (i == 0):
                first = id
 
            # Last Node
            if (i == size - 1):
                last = id
 
            if (temp.left):
                q.append([temp.left, id * 2 + 1])
            if (temp.right):
                q.append([temp.right, id * 2 + 2])
 
        maxWidth = max(maxWidth, last - first + 1)
 
    return maxWidth
 
# Driver Code
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.right = Node(8)
root.right.right.left = Node(6)
root.right.right.right = Node(7)
 
# Function Call
print(getMaxWidth(root))
 
# This code is contributed by Saurabh jaiswal


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
public class GFG {
 
  // Tree Node structure
  public class Node {
    public int data;
    public Node left;
    public Node right;
 
    // Constructor
    public Node(int item) {
      this.data = item;
      this.left = this.right = null;
    }
  };
 
  public class Pair {
    public Node first;
    public int second;
 
    public Pair(Node n, int i) {
      this.first = n;
      this.second = i;
    }
  }
 
  // Function to find the width
  // of the given tree
  static int getMaxWidth(Node root) {
    if (root == null) {
      return 0;
    }
    Queue<Pair> q = new Queue<Pair>();
    q.Enqueue(new Pair(root, 0));
    int maxWidth = 1;
 
    while (q.Count!=0) {
      int size = q.Count;
      int first = 0, last = 0;
 
      for (int i = 0; i < size; i++) {
        Node temp = q.Peek().first;
        int id = q.Peek().second;
        q.Dequeue();
 
        // First Node
        if (i == 0) {
          first = id;
        }
 
        // Last Node
        if (i == size - 1) {
          last = id;
        }
        ;
 
        if (temp.left != null)
          q.Enqueue(new Pair(temp.left, id * 2 + 1));
        if (temp.right != null)
          q.Enqueue(new Pair(temp.right, id * 2 + 2));
      }
 
      maxWidth = Math.Max(maxWidth, last - first + 1);
    }
    return maxWidth;
  }
 
  // Driver Code
  public static void Main(String []args) {
    Node root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.left.right = new Node(5);
    root.right.right = new Node(8);
    root.right.right.left = new Node(6);
    root.right.right.right = new Node(7);
 
    // Function Call
    Console.WriteLine(getMaxWidth(root));
  }
}
 
// This code is contributed by shikhasingrajput


Javascript




<script>
// Javascript program for the above approach
 
// Tree Node structure
class Node {
 
    // Constructor
    constructor(item) {
        this.data = item;
        this.left = this.right = null;
    }
};
 
// Function to find the width
// of the given tree
function getMaxWidth(root) {
    if (root == null) {
        return 0;
    }
    let q = [];
    q.push([root, 0]);
    let maxWidth = 1;
 
    while (q.length) {
        let size = q.length;
        let first, last;
 
        for (let i = 0; i < size; i++) {
            let temp = q[0][0];
            let id = q[0][1];
            q.shift();
 
            // First Node
            if (i == 0) {
                first = id;
            }
 
            // Last Node
            if (i == size - 1) {
                last = id;
            };
 
            if (temp.left)
                q.push([temp.left, id * 2 + 1]);
            if (temp.right)
                q.push([temp.right, id * 2 + 2]);
        }
 
        maxWidth = Math.max(maxWidth, last - first + 1);
    }
    return maxWidth;
}
 
// Driver Code
 
let root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.right = new Node(8);
root.right.right.left = new Node(6);
root.right.right.right = new Node(7);
 
// Function Call
document.write(getMaxWidth(root));
 
// This code is contributed by gfgking.
</script>


 
 

Output

4

 

Time Complexity: O(N)
Auxiliary Space: O(N)

 


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