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Maximum width of an N-ary tree
  • Last Updated : 21 Jun, 2021

Given an N-ary tree, the task is to find the maximum width of the given tree. The maximum width of a tree is the maximum of width among all levels.

Examples:

Input: 

                   4
                 / | \
                2  3 -5
               / \    /\
             -1   3 -2  6

Output:
Explanation: 
Width of 0th level is 1. 
Width of 1st level is 3. 
Width of 2nd level is 4. 
Therefore, the maximum width is 4

Input:  



              1
            / | \
           2  -1  3
         /  \      \
        4    5      8
      /           / | \
     2           6  12  7  

Output:
 

Approach: This problem can be solved using BFS. The idea is to perform level order traversal of the tree. While doing traversal, process nodes of different levels separately. For every level being processed, count the number of nodes present at each level and keep track of the maximum count. Follow the steps below to solve the problem:

  1. Initialize a variable, say maxWidth to store the required maximum width of the tree.
  2. Initialize a Queue to perform the level order traversal of the given tree.
  3. Push the root node into the queue.
  4. If size of the queue exceeds maxWidth for any level, then update maxWidth to the size of the queue.
  5. Traverse the queue and push all the nodes of the next level into the queue and pop all the nodes of the current level.
  6. Repeat the above steps until all the levels of the tree are traversed.
  7. Finally, return the final value of maxWidth.

Below is the implementation of the above approach:

C++




// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum width of
// the tree using level order traversal
int maxWidth(int N, int M,
             vector<int> cost,
             vector<vector<int> > s)
{
    // Store the edges of the tree
    vector<int> adj[N];
    for (int i = 0; i < M; i++) {
        adj[s[i][0]].push_back(
            s[i][1]);
    }
 
    // Stores maximum width
    // of the tree
    int result = 0;
 
    // Stores the nodes
    // of each level
    queue<int> q;
 
    // Insert root node
    q.push(0);
 
    // Perform level order
    // traversal on the tree
    while (!q.empty()) {
 
        // Stores the size of
        // the queue
        int count = q.size();
 
        // Update maximum width
        result = max(count, result);
 
        // Push the nodes of the next
        // level and pop the elements
        // of the current level
        while (count--) {
 
            // Get element from the
            // front the Queue
            int temp = q.front();
            q.pop();
 
            // Push all nodes of the next level.
            for (int i = 0; i < adj[temp].size();
                 i++) {
                q.push(adj[temp][i]);
            }
        }
    }
 
    // Return the result.
    return result;
}
 
// Driver Code
int main()
{
    int N = 11, M = 10;
 
    vector<vector<int> > edges;
    edges.push_back({ 0, 1 });
    edges.push_back({ 0, 2 });
    edges.push_back({ 0, 3 });
    edges.push_back({ 1, 4 });
    edges.push_back({ 1, 5 });
    edges.push_back({ 3, 6 });
    edges.push_back({ 4, 7 });
    edges.push_back({ 6, 10 });
    edges.push_back({ 6, 8 });
    edges.push_back({ 6, 9 });
 
    vector<int> cost
        = { 1, 2, -1, 3, 4, 5,
            8, 2, 6, 12, 7 };
 
    /* Constructed tree is:
                1
              / | \
            2  -1  3
          /  \      \
         4    5      8
        /          / | \
       2          6  12  7
    */
 
    cout << maxWidth(N, M, cost, edges);
 
    return 0;
}


Java




// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
class GFG
{
   
    // Function to find the maximum width of
    // the tree using level order traversal
    static int maxWidth(int N, int M,ArrayList<Integer> cost,
                        ArrayList<ArrayList<Integer> > s)
    {
       
        // Store the edges of the tree
        ArrayList<ArrayList<Integer> > adj =
          new ArrayList<ArrayList<Integer> >();
        for(int i = 0; i < N; i++)
        {
            adj.add(new ArrayList<Integer>());
        }
        for(int i = 0; i < M; i++)
        {
            adj.get(s.get(i).get(0)).add(s.get(i).get(1));
        }
       
        // Stores maximum width
        // of the tree
        int result = 0;
       
        // Stores the nodes
        // of each level
        Queue<Integer> q = new LinkedList<>();
       
        // Insert root node
        q.add(0);
       
        // Perform level order
        // traversal on the tree
        while(q.size() != 0)
        {
           
            // Stores the size of
            // the queue
            int count = q.size();
           
            // Update maximum width
            result = Math.max(count, result);
           
            // Push the nodes of the next
            // level and pop the elements
            // of the current level
            while(count-->0)
            {
               
                // Get element from the
                // front the Queue
                 int temp = q.remove();
               
                // Push all nodes of the next level.
                 for(int i = 0; i < adj.get(temp).size(); i++)
                 {
                     q.add(adj.get(temp).get(i));
                 }
            }
        }
       
        // Return the result.
        return result;
    }
   
    // Driver Code
    public static void main (String[] args)
    {
        int N = 11, M = 10;
        ArrayList<ArrayList<Integer> > edges = new ArrayList<ArrayList<Integer> >();
        edges.add(new ArrayList<Integer>(Arrays.asList( 0, 1)));
        edges.add(new ArrayList<Integer>(Arrays.asList( 0, 2)));
        edges.add(new ArrayList<Integer>(Arrays.asList( 0, 3)));
        edges.add(new ArrayList<Integer>(Arrays.asList(1,4)));
        edges.add(new ArrayList<Integer>(Arrays.asList(1,5)));
        edges.add(new ArrayList<Integer>(Arrays.asList(3,6)));
        edges.add(new ArrayList<Integer>(Arrays.asList(4,7)));
        edges.add(new ArrayList<Integer>(Arrays.asList(6,10)));
        edges.add(new ArrayList<Integer>(Arrays.asList(6,8)));
        edges.add(new ArrayList<Integer>(Arrays.asList(6,9)));
 
        ArrayList<Integer> cost = new ArrayList<Integer>(Arrays.asList(1, 2, -1, 3, 4, 5,8, 2, 6, 12, 7 ));       
        /* Constructed tree is:
                1
              / | \
            2  -1  3
          /  \      \
         4    5      8
        /          / | \
       2          6  12  7
    */
        System.out.println(maxWidth(N, M, cost, edges));
    }
}
 
// This code is contributed by avanitrachhadiya2155


Python3




# Python3 program to implement
# the above approach
from collections import deque
 
# Function to find the maximum width of
#. he tree using level order traversal
def maxWidth(N, M, cost, s):
     
    # Store the edges of the tree
    adj = [[] for i in range(N)]
    for i in range(M):
        adj[s[i][0]].append(s[i][1])
 
    # Stores maximum width
    # of the tree
    result = 0
 
    # Stores the nodes
    # of each level
    q = deque()
 
    # Insert root node
    q.append(0)
 
    # Perform level order
    # traversal on the tree
    while (len(q) > 0):
 
        # Stores the size of
        # the queue
        count = len(q)
 
        # Update maximum width
        result = max(count, result)
 
        # Push the nodes of the next
        # level and pop the elements
        # of the current level
        while (count > 0):
 
            # Get element from the
            # front the Queue
            temp = q.popleft()
 
            # Push all nodes of the next level.
            for i in adj[temp]:
                q.append(i)
                 
            count -= 1
 
    # Return the result.
    return result
 
# Driver Code
if __name__ == '__main__':
     
    N = 11
    M = 10
 
    edges = []
    edges.append([0, 1])
    edges.append([0, 2])
    edges.append([0, 3])
    edges.append([1, 4])
    edges.append([1, 5])
    edges.append([3, 6])
    edges.append([4, 7])
    edges.append([6, 1])
    edges.append([6, 8])
    edges.append([6, 9])
 
    cost = [ 1, 2, -1, 3, 4, 5,
             8, 2, 6, 12, 7]
     
    # Constructed tree is:
    #           1
    #         / | \
    #        2 -1  3
    #       / \       \
    #      4   5        8
    #     /          / | \
    #  2         6  12  7
    print(maxWidth(N, M, cost, edges))
 
# This code is contributed by mohit kumar 29


C#




// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to find the maximum width of
// the tree using level order traversal
static int maxWidth(int N, int M, List<int> cost,
                             List<List<int>> s)
{
     
    // Store the edges of the tree
    List<List<int>> adj = new List<List<int>>();
    for(int i = 0; i < N; i++)
    {
        adj.Add(new List<int>());
    }
     
    for(int i = 0; i < M; i++)
    {
        adj[s[i][0]].Add(s[i][1]);
    }
     
    // Stores maximum width
    // of the tree
    int result = 0;
     
    // Stores the nodes
    // of each level
    Queue<int> q = new Queue<int>();
     
    // Insert root node
    q.Enqueue(0);
     
    // Perform level order
    // traversal on the tree
    while (q.Count != 0)
    {
         
        // Stores the size of
        // the queue
        int count = q.Count;
         
        // Update maximum width
        result = Math.Max(count, result);
         
        // Push the nodes of the next
        // level and pop the elements
        // of the current level
        while (count-- > 0)
        {
             
            // Get element from the
            // front the Queue
            int temp = q.Dequeue();
             
            // Push all nodes of the next level.
            for(int i = 0; i < adj[temp].Count; i++)
            {
                q.Enqueue(adj[temp][i]);
            }
        }
    }
     
    // Return the result.
    return result;
}
 
// Driver Code
static public void Main()
{
    int N = 11, M = 10;
     
    List<List<int>> edges = new List<List<int>>();
    edges.Add(new List<int>(){0, 1});
    edges.Add(new List<int>(){0, 2});
    edges.Add(new List<int>(){0, 3});
    edges.Add(new List<int>(){1, 4});
    edges.Add(new List<int>(){1, 5});
    edges.Add(new List<int>(){3, 6});
    edges.Add(new List<int>(){4, 7});
    edges.Add(new List<int>(){6, 10});
    edges.Add(new List<int>(){6, 8});
    edges.Add(new List<int>(){6, 9});
     
    List<int> cost = new List<int>(){ 1, 2, -1, 3,
                                      4, 5, 8, 2,
                                      6, 12, 7 };
    /* Constructed tree is:
            1
          / | \
        2  -1  3
      /  \      \
     4    5      8
    /          / | \
   2          6  12  7
    */
     
    Console.WriteLine(maxWidth(N, M, cost, edges));
}
}
 
// This code is contributed by rag2127


Javascript




<script>
 
    // JavaScript program for the above approach
     
    // Function to find the maximum width of
    // the tree using level order traversal
    function maxWidth(N, M, cost, s)
    {
        
        // Store the edges of the tree
        let adj = [];
        for(let i = 0; i < N; i++)
        {
            adj.push([]);
        }
        for(let i = 0; i < M; i++)
        {
            adj[s[i][0]].push(s[i][1]);
        }
        
        // Stores maximum width
        // of the tree
        let result = 0;
        
        // Stores the nodes
        // of each level
        let q = [];
        
        // Insert root node
        q.push(0);
        
        // Perform level order
        // traversal on the tree
        while(q.length != 0)
        {
            
            // Stores the size of
            // the queue
            let count = q.length;
            
            // Update maximum width
            result = Math.max(count, result);
            
            // Push the nodes of the next
            // level and pop the elements
            // of the current level
            while(count-->0)
            {
                
                // Get element from the
                // front the Queue
                 let temp = q.shift();
                
                // Push all nodes of the next level.
                 for(let i = 0; i < adj[temp].length; i++)
                 {
                     q.push(adj[temp][i]);
                 }
            }
        }
        
        // Return the result.
        return result;
    }
     
    let N = 11, M = 10;
    let edges = [];
    edges.push([ 0, 1]);
    edges.push([ 0, 2]);
    edges.push([ 0, 3]);
    edges.push([1,4]);
    edges.push([1,5]);
    edges.push([3,6]);
    edges.push([4,7]);
    edges.push([6,10]);
    edges.push([6,8]);
    edges.push([6,9]);
 
    let cost = [1, 2, -1, 3, 4, 5,8, 2, 6, 12, 7 ];      
    /* Constructed tree is:
                  1
                / | \
              2  -1  3
            /  \      \
           4    5      8
          /          / | \
         2          6  12  7
      */
    document.write(maxWidth(N, M, cost, edges));
    
</script>


Output: 

4

 

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

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