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

Recommended : Please try your approach on {IDE} first, before moving on to the solution.

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 :

 // C++ program to find sum of all // elements in generic tree #include using namespace std;    // Represents a node of an n-ary tree struct Node {     int key;     vector 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 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->child).push_back(newNode(15));     (root->child->child).push_back(newNode(20));     (root->child->child).push_back(newNode(30));     (root->child->child).push_back(newNode(40));     (root->child->child).push_back(newNode(100));     (root->child->child).push_back(newNode(20));     (root->child->child->child).push_back(newNode(25));     (root->child->child->child).push_back(newNode(50));        cout << sumNodes(root) << endl;        return 0; }

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.

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