Data Modelling & AI Data Structure & Algorithm

Calculate sum of all nodes present in a level for each level of a Tree

23 June 2025

1

Given a Generic Tree consisting of N nodes (rooted at 0) where each node is associated with a value, the task for each level of the Tree is to find the sum of all node values present at that level of the tree.

Examples:

Input: node_number = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }, node_values = { 2, 3, 4, 4, 7, 6, 2, 3, 9, 1 }

Output:
Sum of level 0 = 2
Sum of level 1 = 7
Sum of level 2 = 14
Sum of level 3 = 18
Explanation :

Nodes on level 0 = {1} with value is 2

Nodes on level 1 = {2, 3} and their respective values are {3, 4}. Sum = 7.

Nodes on level 2 = {4, 5, 8} with values {4, 7, 3} respectively. Sum = 14.

Nodes on level 3 = {6, 7, 9, 10} with values {6, 2, 9, 1} respectively. Sum = 18

Input: node_number = { 1 }, node_values = { 10 }
Output: Sum of level 0 = 10

Approach: Follow the steps below to solve the problem:

Traverse the tree using DFS or BFS
Store the level of this node using this approach.
Then, add the node values to the corresponding level of the node in an array, say sum[ ].
Print the array sum[] showing the sum of all nodes on each level.

Below is the implementation of the above approach :

C++

// C++ implementation of
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to add edges to the tree
void add_edge(int a, int b,
              vector<vector<int> >& tree)
{
    // 0-based indexing
    a--, b--;
 
    tree[a].push_back(b);
    tree[b].push_back(a);
}
 
// Function to print sum of
// nodes on all levels of a tree
void dfs(int u, int level, int par,
         int node_values[], vector<vector<int> >& tree,
         map<int, int>& sum, int& depth)
{
    // update max depth of tree
    depth = max(depth, level);
 
    // Add value of current node
    // to its corresponding level
    sum[level] += node_values[u];
 
    for (int child : tree[u]) {
 
        if (child == par)
            continue;
 
        // Recursive traverse child nodes
        dfs(child, level + 1, u, node_values,
            tree, sum, depth);
    }
}
 
// Function to calculate sum of
// nodes of each level of the Tree
void getSum(int node_values[],
            vector<vector<int> >& tree)
{
    // Depth of the tree
    int depth = 0;
 
    // Stores sum at each level
    map<int, int> sum;
 
    dfs(0, 0,
        -1, node_values,
        tree, sum, depth);
 
    // Print final sum
    for (int i = 0; i <= depth; i++) {
        cout << "Sum of level " << i
             << " = " << sum[i] << endl;
    }
}
 
// Driver Code
int32_t main()
{
 
    // Create a tree structure
    int N = 10;
 
    vector<vector<int> > tree(N);
    add_edge(1, 2, tree);
    add_edge(1, 3, tree);
    add_edge(2, 4, tree);
    add_edge(3, 5, tree);
    add_edge(3, 8, tree);
    add_edge(5, 6, tree);
    add_edge(5, 7, tree);
    add_edge(8, 9, tree);
    add_edge(8, 10, tree);
 
    int node_values[]
        = { 2, 3, 4, 4, 7,
            6, 2, 3, 9, 1 };
 
    // Function call to get the sum
    // of nodes of different level
    getSum(node_values, tree);
 
    return 0;
}

Java

// Java implementation of
// the above approach
import java.io.*;
import java.util.*;
 
class GFG{
     
static Map<Integer, Integer> sum = new HashMap<>();
static int depth = 0;
 
// Function to add edges to the tree
static void add_edge(int a, int b, 
                     ArrayList<ArrayList<Integer>> tree)
{
     
    // 0-based indexing
    a--;
    b--;
  
    tree.get(a).add(b);
    tree.get(b).add(a);
}
  
// Function to print sum of
// Nodes on all levels of a tree
static void dfs(int u, int level, int par,
                int []node_values, 
                ArrayList<ArrayList<Integer>> tree)
{
     
    // Update max depth of tree
    depth = Math.max(depth, level);
  
    // Add value of current node
    // to its corresponding level
    if (sum.containsKey(level))
    {
        sum.put(level, sum.get(level) + 
                       node_values[u]);
    }
    else
        sum.put(level,node_values[u]);
       
    for(int child : tree.get(u))
    {
        if (child == par)
            continue;
  
        // Recursive traverse child nodes
        dfs(child, level + 1, u, node_values,
            tree);
    }
}
  
// Function to calculate sum of
// nodes of each level of the Tree
static void getSum(int []node_values, 
                   ArrayList<ArrayList<Integer>> tree)
{
  
    dfs(0, 0, -1, node_values, tree);
  
    // Print final sum
    for(int i = 0; i <= depth; i++) 
    {
        System.out.println("Sum of level " + (int) i +
                                     " = " + sum.get(i));
    }
}
  
// Driver Code
public static void main (String[] args) 
{
     
    // Create a tree structure
    int N = 10;
  
    ArrayList<ArrayList<Integer>> tree = new ArrayList<ArrayList<Integer>>();
    for(int i = 0; i < N; i++)
       tree.add(new ArrayList<Integer>());
        
    add_edge(1, 2, tree);
    add_edge(1, 3, tree);
    add_edge(2, 4, tree);
    add_edge(3, 5, tree);
    add_edge(3, 8, tree);
    add_edge(5, 6, tree);
    add_edge(5, 7, tree);
    add_edge(8, 9, tree);
    add_edge(8, 10, tree);
  
    int []node_values = { 2, 3, 4, 4, 7,
                          6, 2, 3, 9, 1 };
  
    // Function call to get the sum
    // of nodes of different level
    getSum(node_values, tree);
}
}
 
// This code is contributed by avanitrachhadiya2155

Python3

# Python3 implementation of
# the above approach
 
# Function to add edges to the tree
def add_edge(a, b):
    global tree
     
    # 0-based indexing
    a, b = a - 1, b - 1
    tree[a].append(b)
    tree[b].append(a)
 
# Function to print sum of
# nodes on all levels of a tree
def dfs(u, level, par, node_values):
    global sum, tree, depth
     
    # update max depth of tree
    depth = max(depth, level)
 
    # Add value of current node
    # to its corresponding level
    sum[level] = sum.get(level, 0) + node_values[u]
    for child in tree[u]:
        if (child == par):
            continue
 
        # Recursive traverse child nodes
        dfs(child, level + 1, u, node_values)
 
# Function to calculate sum of
# nodes of each level of the Tree
def getSum(node_values):
    global sum, depth, tree
     
    # Depth of the tree
    # depth = 0
 
    # Stores sum at each level
    # map<int, int> sum
    dfs(0, 0, -1, node_values)
 
    # Prfinal sum
    for i in range(depth + 1):
        print("Sum of level", i, "=", sum[i])
 
# Driver Code
if __name__ == '__main__':
 
    # Create a tree structure
    N = 10
    tree = [[] for i in range(N+1)]
    sum = {}
    depth = 0
    add_edge(1, 2)
    add_edge(1, 3)
    add_edge(2, 4)
    add_edge(3, 5)
    add_edge(3, 8)
    add_edge(5, 6)
    add_edge(5, 7)
    add_edge(8, 9)
    add_edge(8, 10)
    node_values = [2, 3, 4, 4, 7, 6, 2, 3, 9, 1]
 
    # Function call to get the sum
    # of nodes of different level
    getSum(node_values)
 
    # This code is contributed by mohit kumar 29.

C#

// C# implementation of
// the above approach
using System;
using System.Collections.Generic;
class GFG
{
  
static Dictionary<int, int> sum = new Dictionary<int,int>();
  static int depth = 0;
   
// Function to add edges to the tree
static void add_edge(int a, int b, List<List<int>> tree)
{
   
    // 0-based indexing
    a--;
    b--;
 
    tree[a].Add(b);
    tree[b].Add(a);
}
 
// Function to print sum of
// Nodes on all levels of a tree
static void dfs(int u, int level, int par,
         int []node_values, List<List<int>> tree
         )
{
   
    // update max depth of tree
    depth = Math.Max(depth, level);
 
    // Add value of current node
    // to its corresponding level
    if(sum.ContainsKey(level))
      sum[level] += node_values[u];
    else
      sum[level] = node_values[u];
 
    foreach (int child in tree[u]) {
 
        if (child == par)
            continue;
 
        // Recursive traverse child nodes
        dfs(child, level + 1, u, node_values,
            tree);
    }
}
 
// Function to calculate sum of
// nodes of each level of the Tree
static void getSum(int []node_values, List<List<int>> tree)
{
 
    dfs(0, 0, -1, node_values, tree);
 
    // Print final sum
    for (int i = 0; i <= depth; i++) {
        Console.WriteLine("Sum of level " + (int) i + " = "+ sum[i]);
    }
}
 
// Driver Code
public static void Main()
{
 
    // Create a tree structure
    int N = 10;
 
    List<List<int> > tree = new List<List<int>>();
    for(int i = 0; i < N; i++)
       tree.Add(new List<int>());
    add_edge(1, 2, tree);
    add_edge(1, 3, tree);
    add_edge(2, 4, tree);
    add_edge(3, 5, tree);
    add_edge(3, 8, tree);
    add_edge(5, 6, tree);
    add_edge(5, 7, tree);
    add_edge(8, 9, tree);
    add_edge(8, 10, tree);
 
    int []node_values = {2, 3, 4, 4, 7,6, 2, 3, 9, 1};
 
    // Function call to get the sum
    // of nodes of different level
    getSum(node_values, tree);
}
}
 
// This code is contributed by bgangwar59.

Javascript

<script>
 
// Javascript implementation of
// the above approach
var sum = new Map();
var depth = 0;
   
// Function to add edges to the tree
function add_edge(a, b, tree)
{
     
    // 0-based indexing
    a--;
    b--;
 
    tree[a].push(b);
    tree[b].push(a);
}
 
// Function to print sum of
// Nodes on all levels of a tree
function dfs(u, level, par, node_values, tree)
{
     
    // Update max depth of tree
    depth = Math.max(depth, level);
 
    // Push value of current node
    // to its corresponding level
    if (sum.has(level))
        sum.set(level, sum.get(level) +
                       node_values[u]);
    else
        sum.set(level, node_values[u])
 
    for(var child of tree[u])
    {
        if (child == par)
            continue;
 
        // Recursive traverse child nodes
        dfs(child, level + 1, u, node_values,
            tree);
    }
}
 
// Function to calculate sum of
// nodes of each level of the Tree
function getSum(node_values, tree)
{
    dfs(0, 0, -1, node_values, tree);
 
    // Print final sum
    for(var i = 0; i <= depth; i++) 
    {
        document.write("Sum of level " + i +
                       " = "+ sum.get(i) + "<br>");
    }
}
 
// Driver Code
 
// Create a tree structure
var N = 10;
var tree = [];
for(var i = 0; i < N; i++)
   tree.push([]);
    
add_edge(1, 2, tree);
add_edge(1, 3, tree);
add_edge(2, 4, tree);
add_edge(3, 5, tree);
add_edge(3, 8, tree);
add_edge(5, 6, tree);
add_edge(5, 7, tree);
add_edge(8, 9, tree);
add_edge(8, 10, tree);
var node_values = [ 2, 3, 4, 4, 7,6, 2, 3, 9, 1 ];
 
// Function call to get the sum
// of nodes of different level
getSum(node_values, tree);
 
// This code is contributed by rrrtnx
 
</script>

Output

Sum of level 0 = 2
Sum of level 1 = 7
Sum of level 2 = 14
Sum of level 3 = 18

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

Iterative Approach(Level Order Traversal using Queue Data Structure):
Follow the below steps to solve the above problem:
1) Perform level Order Traversal and keep track of level and sum at each level.
2) At each level calculate sum and print sum along with level.
3) Repeat the step-2 at each level till last level.

Below is the implementation of above approach:

C++

// THIS CODE IS CONTRIBUTED BY KIRTI AGARWAL(KIRTIAGARWAL23121999)
// C++ Implementation of above approach
#include<bits/stdc++.h>
using namespace std;
 
// a binary tree node
struct Node{
    int data;
    Node *left, *right;
    Node(int data){
        this->data = data;
        this->left = this->right = NULL;
    }
};
 
// a utility function to create a new node
Node* newNode(int data){
    return new Node(data);
}
 
void getSum(int node_values[], Node* root){
    queue<Node*> q;
    q.push(root);
    int level = 0;
    while(!q.empty()){
        int n = q.size();
        int sum = 0;
        for(int i = 0; i<n; i++){
            Node* front_node = q.front();
            q.pop();
            sum += node_values[front_node->data - 1];
            if(front_node->left) q.push(front_node->left);
            if(front_node->right) q.push(front_node->right);
        }
        cout<<"Sum of level "<<level<<" : "<<sum<<endl;
        level++;
    }
}
 
// driver code to test above function
int main(){
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->right->left = newNode(5);
    root->right->right = newNode(8);
    root->right->left->left = newNode(6);
    root->right->left->right = newNode(7);
    root->right->right->left = newNode(9);
    root->right->right->right = newNode(10);
     
    int node_values[] = {2,3,4,4,7,6,2,3,9,1};
     
    // function call to get the sum
    // of nodes of different level
    getSum(node_values, root);
    return 0;
}

Java

import java.util.LinkedList;
import java.util.Queue;
 
class Node {
    int data;
    Node left, right;
 
    public Node(int data)
    {
        this.data = data;
        this.left = this.right = null;
    }
}
 
public class GFG {
    public static void getSum(int[] node_values, Node root)
    {
       
        // Create a queue to store the nodes of the binary
        // tree
        Queue<Node> q = new LinkedList<>();
       
        // Add the root node to the queue
        q.add(root);
       
        // Initialize the level to 0
        int level = 0;
       
        // Loop until the queue is empty
        while (!q.isEmpty()) 
        {
           
            // Get the number of nodes at the current level
            int n = q.size();
           
            // Initialize the sum to 0
            int sum = 0;
           
            // Loop through all the nodes at the current
            // level
            for (int i = 0; i < n; i++) 
            {
               
                // Get the front node from the queue
                Node front_node = q.poll();
               
                // Add the value of the node to the sum
                sum += node_values[front_node.data - 1];
               
                // Add the left child of the front node to
                // the queue if it exists
                if (front_node.left != null) {
                    q.add(front_node.left);
                }
               
                // Add the right child of the front node to
                // the queue if it exists
                if (front_node.right != null) {
                    q.add(front_node.right);
                }
            }
           
            // Print the sum of the nodes at the current
            // level
            System.out.println("Sum of level " + level
                               + " : " + sum);
            // Increment the level
            level++;
        }
    }
 
    public static void main(String[] args)
    {
        // Create the binary tree
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.right.left = new Node(5);
        root.right.right = new Node(8);
        root.right.left.left = new Node(6);
        root.right.left.right = new Node(7);
        root.right.right.left = new Node(9);
        root.right.right.right = new Node(10);
 
        // Define the values of the nodes
        int[] node_values
            = { 2, 3, 4, 4, 7, 6, 2, 3, 9, 1 };
 
        // Call the function to get the sum of nodes at
        // different levels
        getSum(node_values, root);
    }
}

Python3

# a binary tree node
class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
def newNode(data):
    return Node(data)
 
def getSum(node_values, root):
    q = []
    q.append(root)
    level = 0
    while len(q) != 0:
        n = len(q)
        total = 0
        for i in range(n):
            front_node = q.pop(0)
            total += node_values[front_node.data - 1]
            if front_node.left:
                q.append(front_node.left)
            if front_node.right:
                q.append(front_node.right)
        print(f"Sum of level {level} : {total}")
        level += 1
 
# driver code to test above function
root = newNode(1)
root.left = newNode(2)
root.right = newNode(3)
root.left.left = newNode(4)
root.right.left = newNode(5)
root.right.right = newNode(8)
root.right.left.left = newNode(6)
root.right.left.right = newNode(7)
root.right.right.left = newNode(9)
root.right.right.right = newNode(10)
 
node_values = [2, 3, 4, 4, 7, 6, 2, 3, 9, 1]
 
# function call to get the sum
# of nodes of different level
getSum(node_values, root)

Javascript

// a binary tree node
class Node {
  constructor(data) {
    this.data = data;
    this.left = null;
    this.right = null;
  }
}
 
function newNode(data) {
  return new Node(data);
}
 
function getSum(node_values, root) {
  let q = [];
  q.push(root);
  let level = 0;
  while(q.length !== 0){
    let n = q.length;
    let sum = 0;
    for(let i = 0; i < n; i++){
      let front_node = q.shift();
      sum += node_values[front_node.data - 1];
      if(front_node.left) q.push(front_node.left);
      if(front_node.right) q.push(front_node.right);
    }
    console.log(`Sum of level ${level} : ${sum}`);
    level++;
  }
}
 
// driver code to test above function
let root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.right.left = newNode(5);
root.right.right = newNode(8);
root.right.left.left = newNode(6);
root.right.left.right = newNode(7);
root.right.right.left = newNode(9);
root.right.right.right = newNode(10);
 
let node_values = [2, 3, 4, 4, 7, 6, 2, 3, 9, 1];
 
// function call to get the sum
// of nodes of different level
getSum(node_values, root);

C#

using System;
using System.Collections.Generic;
 
public class Node{
    public int data;
    public Node left, right;
    public Node(int data){
        this.data = data;
        this.left = this.right = null;
    }
};
 
public class Gfg{
    public static Node newNode(int data){
        return new Node(data);
    }
 
    public static void getSum(int[] node_values, Node root){
        Queue<Node> q = new Queue<Node>();
        q.Enqueue(root);
        int level = 0;
        while(q.Count != 0){
            int n = q.Count;
            int sum = 0;
            for(int i = 0; i<n; i++){
                Node front_node = q.Peek();
                q.Dequeue();
                sum += node_values[front_node.data - 1];
                if(front_node.left != null) q.Enqueue(front_node.left);
                if(front_node.right != null) q.Enqueue(front_node.right);
            }
            Console.WriteLine("Sum of level "+level+" : "+sum);
            level++;
        }
    }
 
    public static void Main(){
        Node root = newNode(1);
        root.left = newNode(2);
        root.right = newNode(3);
        root.left.left = newNode(4);
        root.right.left = newNode(5);
        root.right.right = newNode(8);
        root.right.left.left = newNode(6);
        root.right.left.right = newNode(7);
        root.right.right.left = newNode(9);
        root.right.right.right = newNode(10);
     
        int[] node_values = new int[]{2,3,4,4,7,6,2,3,9,1};
     
        // function call to get the sum
        // of nodes of different level
        getSum(node_values, root);
    }
}

Output

Sum of level 0 : 2
Sum of level 1 : 7
Sum of level 2 : 14
Sum of level 3 : 18

Time Complexity: O(N) where N is the number of nodes in given Binary Tree.
Auxiliary Space: O(N) due to queue data structure.

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Calculate sum of all nodes present in a level for each level of a Tree

C++

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C#

Javascript

C++

Java

Python3

Javascript

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Samsung’s June update brings Galaxy S25 three powerful AI tricks the S26 already had

Your Galaxy S25 just picked up three more Galaxy AI features

No one can agree on these phones. [Video]

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Samsung’s June update brings Galaxy S25 three powerful AI tricks the S26 already had

Your Galaxy S25 just picked up three more Galaxy AI features

No one can agree on these phones. [Video]

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