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 :
C++
// 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; } |
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Java
// Java program to find sum of all // elements in generic tree import java.util.*; class GFG { // Represents a node of an n-ary tree static class Node { int key; Vector<Node> child; }; // Utility function to create a new tree node static Node newNode(int key) { Node temp = new Node(); temp.key = key; temp.child = new Vector<>(); return temp; } // Function to compute the sum // of all elements in generic tree static 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 = new LinkedList<>(); q.add(root); while (!q.isEmpty()) { 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.peek(); q.remove(); sum += p.key; // Enqueue all children of the dequeued item for (int i = 0; i < p.child.size(); i++) q.add(p.child.get(i)); n--; } } return sum; } // Driver program public static void main(String[] args) { // Creating a generic tree Node root = newNode(20); (root.child).add(newNode(2)); (root.child).add(newNode(34)); (root.child).add(newNode(50)); (root.child).add(newNode(60)); (root.child).add(newNode(70)); (root.child.get(0).child).add(newNode(15)); (root.child.get(0).child).add(newNode(20)); (root.child.get(1).child).add(newNode(30)); (root.child.get(2).child).add(newNode(40)); (root.child.get(2).child).add(newNode(100)); (root.child.get(2).child).add(newNode(20)); (root.child.get(0).child.get(1).child).add(newNode(25)); (root.child.get(0).child.get(1).child).add(newNode(50)); System.out.print(sumNodes(root) +"\n"); } } // This code is contributed by 29AjayKumar |
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C#
// C# program to find sum of all // elements in generic tree using System; using System.Collections.Generic; class GFG { // Represents a node of an n-ary tree class Node { public int key; public List<Node> child; }; // Utility function to create a new tree node static Node newNode(int key) { Node temp = new Node(); temp.key = key; temp.child = new List<Node>(); return temp; } // Function to compute the sum // of all elements in generic tree static 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 = new Queue<Node>(); q.Enqueue(root); while (q.Count != 0) { int n = q.Count; // If this node has children while (n > 0) { // Dequeue an item from queue and // add it to variable "sum" Node p = q.Peek(); q.Dequeue(); sum += p.key; // Enqueue all children of the dequeued item for (int i = 0; i < p.child.Count; i++) q.Enqueue(p.child[i]); n--; } } return sum; } // Driver program public static void Main(String[] args) { // Creating a generic tree Node root = newNode(20); (root.child).Add(newNode(2)); (root.child).Add(newNode(34)); (root.child).Add(newNode(50)); (root.child).Add(newNode(60)); (root.child).Add(newNode(70)); (root.child[0].child).Add(newNode(15)); (root.child[0].child).Add(newNode(20)); (root.child[1].child).Add(newNode(30)); (root.child[2].child).Add(newNode(40)); (root.child[2].child).Add(newNode(100)); (root.child[2].child).Add(newNode(20)); (root.child[0].child[1].child).Add(newNode(25)); (root.child[0].child[1].child).Add(newNode(50)); Console.Write(sumNodes(root) +"\n"); } } // This code is contributed by PrinciRaj1992 |
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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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