Sum of all elements of N-ary Tree

Given an N-ary tree, find sum of all elements in it.

Example : 



Input : Above tree
Output : Sum is 536

Approach : The approach used is similar to Level Order traversal in a binary tree. Start by pushing the root node in the queue. And for each node, while popping it from queue, add the value of this node in the sum variable and push the children of the popped element in the queue. In case of a generic tree store child nodes in a vector. Thus, put all elements of the vector in the queue.

Below is the implementation of the above idea :

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find sum of all
// elements in generic tree
#include <bits/stdc++.h>
using namespace std;
  
// Represents a node of an n-ary tree
struct Node {
    int key;
    vector<Node*> child;
};
  
// Utility function to create a new tree node
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    return temp;
}
  
// Function to compute the sum
// of all elements in generic tree
int sumNodes(Node* root)
{
    // initialize the sum variable
    int sum = 0;
  
    if (root == NULL)
        return 0;
  
    // Creating a queue and pushing the root
    queue<Node*> q;
    q.push(root);
  
    while (!q.empty()) {
        int n = q.size();
  
        // If this node has children
        while (n > 0) {
  
            // Dequeue an item from queue and
            // add it to variable "sum"
            Node* p = q.front();
            q.pop();
            sum += p->key;
  
            // Enqueue all children of the dequeued item
            for (int i = 0; i < p->child.size(); i++)
                q.push(p->child[i]);
            n--;
        }
    }
    return sum;
}
  
// Driver program
int main()
{
    // Creating a generic tree
    Node* root = newNode(20);
    (root->child).push_back(newNode(2));
    (root->child).push_back(newNode(34));
    (root->child).push_back(newNode(50));
    (root->child).push_back(newNode(60));
    (root->child).push_back(newNode(70));
    (root->child[0]->child).push_back(newNode(15));
    (root->child[0]->child).push_back(newNode(20));
    (root->child[1]->child).push_back(newNode(30));
    (root->child[2]->child).push_back(newNode(40));
    (root->child[2]->child).push_back(newNode(100));
    (root->child[2]->child).push_back(newNode(20));
    (root->child[0]->child[1]->child).push_back(newNode(25));
    (root->child[0]->child[1]->child).push_back(newNode(50));
  
    cout << sumNodes(root) << endl;
  
    return 0;
}

chevron_right






                        filter_none








Output:


536



Time Complexity : O(N), where N is the number of nodes in tree.

Auxiliary Space : O(N), where N is the number of nodes in tree.








          
          
          
          

            

    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
Sahil_Bansall
In love with a semicolon because sometimes i miss it so badly)
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.