# 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.

This article is attributed to GeeksforGeeks.org

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