Data Modelling & AI Data Structure & Algorithm

Maximum level sum in N-ary Tree

23 June 2025

0

Given an N-ary Tree consisting of nodes valued [1, N] and an array value[], where each node i is associated with value[i], the task is to find the maximum of sum of all node values of all levels of the N-ary Tree.

Examples:

Input: N = 8, Edges[][2] = {{0, 1}, {0, 2}, {0, 3}, {1, 4}, {1, 5}, {3, 6}, {6, 7}}, Value[] = {4, 2, 3, -5, -1, 3, -2, 6}
Output: 6
Explanation:
Sum of all nodes of 0^th level is 4
Sum of all nodes of 1^st level is 0
Sum of all the nodes of 3^rd level is 0.
Sum of all the nodes of 4^th level is 6.
Therefore, maximum sum of any level of the tree is 6.

Input: N = 10, Edges[][2] = {{0, 1}, {0, 2}, {0, 3}, {1, 4}, {1, 5}, {3, 6}, {6, 7}, {6, 8}, {6, 9}}, Value[] = {1, 2, -1, 3, 4, 5, 8, 6, 12, 7}
Output: 25

Approach: This problem can be solved using Level order Traversal. While performing the traversal, process nodes of different levels separately. For every level being processed, compute the sum of nodes at that level and keep track of the maximum sum. Follow the steps:

Store all the child nodes at the current level in the queue and then count the total sum of nodes at the current level after the level order traversal for a particular level is completed.
Since the queue now contains all the nodes of the next level, the total sum of nodes in the next level can be easily calculated by traversing the queue.
Follow the same procedure for the successive levels and update the maximum sum of nodes found at each level.
After the above steps, print the maximum sum of values stored.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum sum
// a level in N-ary treeusing BFS
int maxLevelSum(int N, int M,
                vector<int> Value,
                int Edges[][2])
{
    // Stores the edges of the graph
    vector<int> adj[N];
 
    // Create Adjacency list
    for (int i = 0; i < M; i++) {
        adj[Edges[i][0]].push_back(
            Edges[i][1]);
    }
 
    // Initialize result
    int result = Value[0];
 
    // Stores the nodes of each level
    queue<int> q;
 
    // Insert root
    q.push(0);
 
    // Perform level order traversal
    while (!q.empty()) {
 
        // Count of nodes of the
        // current level
        int count = q.size();
 
        int sum = 0;
 
        // Traverse the current level
        while (count--) {
 
            // Dequeue a node from queue
            int temp = q.front();
            q.pop();
 
            // Update sum of current level
            sum = sum + Value[temp];
 
            // Enqueue the children of
            // dequeued node
            for (int i = 0;
                 i < adj[temp].size(); i++) {
                q.push(adj[temp][i]);
            }
        }
 
        // Update maximum level sum
        result = max(sum, result);
    }
 
    // Return the result
    return result;
}
 
// Driver Code
int main()
{
    // Number of nodes
    int N = 10;
 
    // Edges of the N-ary tree
    int Edges[][2] = { { 0, 1 }, { 0, 2 },
                       { 0, 3 }, { 1, 4 },
                       { 1, 5 }, { 3, 6 },
                       { 6, 7 }, { 6, 8 },
                       { 6, 9 } };
    // Given cost
    vector<int> Value = { 1, 2, -1, 3, 4,
                          5, 8, 6, 12, 7 };
 
    // Function call
    cout << maxLevelSum(N, N - 1,
                        Value, Edges);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
 
class GFG{
 
// Function to find the maximum sum
// a level in N-ary treeusing BFS
@SuppressWarnings("unchecked")
static int maxLevelSum(int N, int M, int[] Value, 
                                     int Edges[][])
{
     
    // Stores the edges of the graph
    ArrayList<Integer>[] adj = new ArrayList[N];
 
    for(int i = 0; i < N; i++)
    {
        adj[i] = new ArrayList<>();
    }
 
    // Create Adjacency list
    for(int i = 0; i < M; i++)
    {
        adj[Edges[i][0]].add(Edges[i][1]);
    }
 
    // Initialize result
    int result = Value[0];
 
    // Stores the nodes of each level
    Queue<Integer> q = new LinkedList<>();
 
    // Insert root
    q.add(0);
 
    // Perform level order traversal
    while (!q.isEmpty()) 
    {
         
        // Count of nodes of the
        // current level
        int count = q.size();
 
        int sum = 0;
 
        // Traverse the current level
        while (count-- > 0) 
        {
             
            // Dequeue a node from queue
            int temp = q.poll();
 
            // Update sum of current level
            sum = sum + Value[temp];
 
            // Enqueue the children of
            // dequeued node
            for(int i = 0; i < adj[temp].size(); i++) 
            {
                q.add(adj[temp].get(i));
            }
        }
 
        // Update maximum level sum
        result = Math.max(sum, result);
    }
 
    // Return the result
    return result;
}
 
// Driver Code
public static void main(String[] args) 
{
     
    // Number of nodes
    int N = 10;
 
    // Edges of the N-ary tree
    int[][] Edges = { { 0, 1 }, { 0, 2 },
                      { 0, 3 }, { 1, 4 },
                      { 1, 5 }, { 3, 6 }, 
                      { 6, 7 }, { 6, 8 }, 
                      { 6, 9 } };
    // Given cost
    int[] Value = { 1, 2, -1, 3, 4,
                    5, 8, 6, 12, 7 };
 
    // Function call
    System.out.println(maxLevelSum(N, N - 1,
                                   Value, Edges));
}
}
 
// This code is contributed by sanjeev2552

Python3

# Python3 program for the above approach
from collections import deque
 
# Function to find the maximum sum
# a level in N-ary treeusing BFS
def maxLevelSum(N, M, Value, Edges):
     
    # Stores the edges of the graph
    adj = [[] for i in range(N)]
 
    # Create Adjacency list
    for i in range(M):
        adj[Edges[i][0]].append(Edges[i][1])
 
    # Initialize result
    result = Value[0]
 
    # Stores the nodes of each level
    q = deque()
 
    # Insert root
    q.append(0)
 
    # Perform level order traversal
    while (len(q) > 0):
 
        # Count of nodes of the
        # current level
        count = len(q)
 
        sum = 0
 
        # Traverse the current level
        while (count):
 
            # Dequeue a node from queue
            temp = q.popleft()
 
            # Update sum of current level
            sum = sum + Value[temp]
 
            # Enqueue the children of
            # dequeued node
            for i in range(len(adj[temp])):
                q.append(adj[temp][i])
                 
            count -= 1
 
        # Update maximum level sum
        result = max(sum, result)
 
    # Return the result
    return result
 
# Driver Code
if __name__ == '__main__':
     
    # Number of nodes
    N = 10
 
    # Edges of the N-ary tree
    Edges = [ [ 0, 1 ], [ 0, 2 ],
              [ 0, 3 ], [ 1, 4 ],
              [ 1, 5 ], [ 3, 6 ],
              [ 6, 7 ], [ 6, 8 ],
              [ 6, 9 ] ]
               
    # Given cost
    Value = [ 1, 2, -1, 3, 4, 
              5, 8, 6, 12, 7 ]
 
    # Function call
    print(maxLevelSum(N, N - 1, 
                      Value, Edges))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the 
// above approach
using System;
using System.Collections.Generic;
class GFG{
 
// Function to find the maximum sum
// a level in N-ary treeusing BFS
 
static int maxLevelSum(int N, int M, 
                       int[] Value, 
                       int [,]Edges)
{
  // Stores the edges of the graph
  List<int>[] adj = new List<int>[N];
 
  for(int i = 0; i < N; i++)
  {
    adj[i] = new List<int>();
  }
 
  // Create Adjacency list
  for(int i = 0; i < M; i++)
  {
    adj[Edges[i, 0]].Add(Edges[i, 1]);
  }
 
  // Initialize result
  int result = Value[0];
 
  // Stores the nodes of each level
  Queue<int> q = new Queue<int>();
 
  // Insert root
  q.Enqueue(0);
 
  // Perform level order 
  // traversal
  while (q.Count != 0) 
  {
    // Count of nodes of the
    // current level
    int count = q.Count;
 
    int sum = 0;
 
    // Traverse the current 
    // level
    while (count-- > 0) 
    {
      // Dequeue a node from 
      // queue
      int temp = q.Peek();
      q.Dequeue();
 
      // Update sum of current 
      // level
      sum = sum + Value[temp];
 
      // Enqueue the children of
      // dequeued node
      for(int i = 0; 
              i < adj[temp].Count; i++) 
      {
        q.Enqueue(adj[temp][i]);
      }
    }
 
    // Update maximum level sum
    result = Math.Max(sum, result);
  }
 
  // Return the result
  return result;
}
 
// Driver Code
public static void Main(String[] args) 
{    
  // Number of nodes
  int N = 10;
 
  // Edges of the N-ary tree
  int[,] Edges = {{0, 1}, {0, 2},
                  {0, 3}, {1, 4},
                  {1, 5}, {3, 6}, 
                  {6, 7}, {6, 8}, 
                  {6, 9}};
  // Given cost
  int[] Value = {1, 2, -1, 3, 4,
                 5, 8, 6, 12, 7};
 
  // Function call
  Console.WriteLine(maxLevelSum(N, N - 1,
                                Value, Edges));
}
}
 
// This code is contributed by Princi Singh

Javascript

<script>
 
    // JavaScript program for the above approach
     
    // Function to find the maximum sum
    // a level in N-ary treeusing BFS
 
    function maxLevelSum(N, M, Value, Edges)
    {
      // Stores the edges of the graph
      let adj = new Array(N);
 
      for(let i = 0; i < N; i++)
      {
        adj[i] = [];
      }
 
      // Create Adjacency list
      for(let i = 0; i < M; i++)
      {
        adj[Edges[i][0]].push(Edges[i][1]);
      }
 
      // Initialize result
      let result = Value[0];
 
      // Stores the nodes of each level
      let q = [];
 
      // Insert root
      q.push(0);
 
      // Perform level order
      // traversal
      while (q.length != 0)
      {
        // Count of nodes of the
        // current level
        let count = q.length;
 
        let sum = 0;
 
        // Traverse the current
        // level
        while (count-- > 0)
        {
          // Dequeue a node from
          // queue
          let temp = q[0];
          q.shift();
 
          // Update sum of current
          // level
          sum = sum + Value[temp];
 
          // Enqueue the children of
          // dequeued node
          for(let i = 0; i < adj[temp].length; i++)
          {
            q.push(adj[temp][i]);
          }
        }
 
        // Update maximum level sum
        result = Math.max(sum, result);
      }
 
      // Return the result
      return result;
    }
     
    // Number of nodes
    let N = 10;
 
    // Edges of the N-ary tree
    let Edges = [[0, 1], [0, 2],
                    [0, 3], [1, 4],
                    [1, 5], [3, 6],
                    [6, 7], [6, 8],
                    [6, 9]];
    // Given cost
    let Value = [1, 2, -1, 3, 4, 5, 8, 6, 12, 7];
 
    // Function call
    document.write(maxLevelSum(N, N - 1, Value, Edges));
 
</script>

Output

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

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1 COMMENT

yarmok 5 December 2025 At 10:33 pm

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Maximum level sum in N-ary Tree

C++

Java

Python3

C#

Javascript

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