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Sum of all nodes in a binary tree

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  • Difficulty Level : Easy
  • Last Updated : 03 Sep, 2021

Give an algorithm for finding the sum of all elements in a binary tree.
 

In the above binary tree sum = 106. 
 

The idea is to recursively, call left subtree sum, right subtree sum and add their values to current node’s data. 
 

C++




/* Program to print sum of all the elements of a binary tree */
#include <bits/stdc++.h>
using namespace std;
 
struct Node {
    int key;
    Node* left, *right;
};
 
/* utility that allocates a new Node with the given key  */
Node* newNode(int key)
{
    Node* node = new Node;
    node->key = key;
    node->left = node->right = NULL;
    return (node);
}
 
/* Function to find sum of all the elements*/
int addBT(Node* root)
{
    if (root == NULL)
        return 0;
    return (root->key + addBT(root->left) + addBT(root->right));
}
 
/* Driver program to test above functions*/
int main()
{
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->right->left->right = newNode(8);
 
    int sum = addBT(root);
    cout << "Sum of all the elements is: " << sum << endl;
 
    return 0;
}


Java




// Java Program to print sum of
// all the elements of a binary tree
class GFG
{
static class Node
{
    int key;
    Node left, right;
}
 
/* utility that allocates a new
   Node with the given key */
static Node newNode(int key)
{
    Node node = new Node();
    node.key = key;
    node.left = node.right = null;
    return (node);
}
 
/* Function to find sum
   of all the elements*/
static int addBT(Node root)
{
    if (root == null)
        return 0;
    return (root.key + addBT(root.left) +
                       addBT(root.right));
}
 
// Driver Code
public static void main(String args[])
{
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.right.left.right = newNode(8);
 
    int sum = addBT(root);
    System.out.println("Sum of all the elements is: " + sum);
}
}
 
// This code is contributed by Arnab Kundu


Python3




# Python3 Program to print sum of all
# the elements of a binary tree
 
# Binary Tree Node
 
""" utility that allocates a new Node
with the given key """
class newNode:
 
    # Construct to create a new node
    def __init__(self, key):
        self.key = key
        self.left = None
        self.right = None
         
# Function to find sum of all the element
def addBT(root):
    if (root == None):
        return 0
    return (root.key + addBT(root.left) +
                       addBT(root.right))
 
# Driver Code
if __name__ == '__main__':
    root = newNode(1)
    root.left = newNode(2)
    root.right = newNode(3)
    root.left.left = newNode(4)
    root.left.right = newNode(5)
    root.right.left = newNode(6)
    root.right.right = newNode(7)
    root.right.left.right = newNode(8)
 
    sum = addBT(root)
 
    print("Sum of all the nodes is:", sum)
 
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)


C#




using System;
 
// C# Program to print sum of
// all the elements of a binary tree
public class GFG
{
public class Node
{
    public int key;
    public Node left, right;
}
 
/* utility that allocates a new 
   Node with the given key */
public static Node newNode(int key)
{
    Node node = new Node();
    node.key = key;
    node.left = node.right = null;
    return (node);
}
 
/* Function to find sum 
   of all the elements*/
public static int addBT(Node root)
{
    if (root == null)
    {
        return 0;
    }
    return (root.key + addBT(root.left) + addBT(root.right));
}
 
// Driver Code
public static void Main(string[] args)
{
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.right.left.right = newNode(8);
 
    int sum = addBT(root);
    Console.WriteLine("Sum of all the elements is: " + sum);
}
}
 
// This code is contributed by Shrikant13


Javascript




<script>
// Javascript Program to print sum of
// all the elements of a binary tree
 
class Node
{   
    constructor(key)
    {
        this.key=key;
        this.left=this.right=null;
    }
}
 
/* Function to find sum
   of all the elements*/
function addBT(root)
{
    if (root == null)
        return 0;
    return (root.key + addBT(root.left) +
                       addBT(root.right));
}
 
// 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.left = new Node(6);
root.right.right = new Node(7);
root.right.left.right = new Node(8);
 
let sum = addBT(root);
document.write("Sum of all the elements is: " + sum);
 
// This code is contributed by avanitrachhadiya2155
</script>


Output

Sum of all the elements is: 36

Method 2 – Another way to solve this problem is by using Level Order Traversal. Every time when a Node is deleted from the queue, add it to the sum variable.

C++




#include <bits/stdc++.h>
#include <iostream>
using namespace std;
 
struct Node {
    int key;
    struct Node *left, *right;
};
 
// Utility function to create a new node
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    temp->left = temp->right = NULL;
    return (temp);
}
 
/*Function to find sum of all elements*/
int sumBT(Node* root)
{
      //sum variable to track the sum of
      //all variables.
    int sum = 0;
   
    queue<Node*> q;
 
      //Pushing the first level.
    q.push(root);
 
      //Pushing elements at each level from
      //the tree.
    while (!q.empty()) {
        Node* temp = q.front();
        q.pop();
       
          //After popping each element from queue
          //add its data to the sum variable.
        sum += temp->key;
 
        if (temp->left) {
            q.push(temp->left);
        }
        if (temp->right) {
            q.push(temp->right);
        }
    }
    return sum;
}
 
// Driver program
int main()
{
    // Let us create Binary Tree shown in above example
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->right->left->right = newNode(8);
 
    cout << "Sum of all elements in the binary tree is: "
         << sumBT(root);
}
 
//This code is contributed by Sarthak Delori


Output

Sum of all elements in the binary tree is: 36

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

This article is contributed by Prakriti Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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