Check if the given n-ary tree is a binary tree
Given an n-ary tree, the task is to check whether the given tree is binary or not.
Examples:
Input:
A
/ \
B C
/ \ \
D E F
Output: Yes
Input:
A
/ | \
B C D
\
F
Output: No
Approach: Every node in a binary tree can have at most 2 children. So, for every node of the given tree, count the number of children and if for any node the count exceeds 2 then print No else print Yes.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Structure of a node// of an n-ary treestruct Node { char key; vector<Node*> child;}; // Utility function to create// a new tree nodeNode* newNode(int key){ Node* temp = new Node; temp->key = key; return temp;} // Function that returns true// if the given tree is binarybool isBinaryTree(struct Node* root){ // Base case if (!root) return true; // Count will store the number of // children of the current node int count = 0; for (int i = 0; i < root->child.size(); i++) { // If any child of the current node doesn't // satify the condition of being // a binary tree node if (!isBinaryTree(root->child[i])) return false; // Increment the count of children count++; // If current node has more // than 2 children if (count > 2) return false; } return true;} // Driver codeint main(){ Node* root = newNode('A'); (root->child).push_back(newNode('B')); (root->child).push_back(newNode('C')); (root->child[0]->child).push_back(newNode('D')); (root->child[0]->child).push_back(newNode('E')); (root->child[1]->child).push_back(newNode('F')); if (isBinaryTree(root)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG { // Structure of a node// of an n-ary treestatic class Node { int key; Vector<Node> child = new Vector<Node>();}; // Utility function to create// a new tree nodestatic Node newNode(int key){ Node temp = new Node(); temp.key = key; return temp;} // Function that returns true// if the given tree is binarystatic boolean isBinaryTree(Node root){ // Base case if (root == null) return true; // Count will store the number of // children of the current node int count = 0; for (int i = 0; i < root.child.size(); i++) { // If any child of the current node doesn't // satify the condition of being // a binary tree node if (!isBinaryTree(root.child.get(i))) return false; // Increment the count of children count++; // If current node has more // than 2 children if (count > 2) return false; } return true;} // Driver codepublic static void main(String[] args) { Node root = newNode('A'); (root.child).add(newNode('B')); (root.child).add(newNode('C')); (root.child.get(0).child).add(newNode('D')); (root.child.get(0).child).add(newNode('E')); (root.child.get(1).child).add(newNode('F')); if (isBinaryTree(root)) System.out.println("Yes"); else System.out.println("No");}} // This code is contributed by PrinciRaj1992 |
C#
// C# implementation of the above approachusing System;using System.Collections.Generic; class GFG { // Structure of a node// of an n-ary treepublic class Node { public int key; public List<Node> child = new List<Node>();}; // Utility function to create// a new tree nodestatic Node newNode(int key){ Node temp = new Node(); temp.key = key; return temp;} // Function that returns true// if the given tree is binarystatic bool isBinaryTree(Node root){ // Base case if (root == null) return true; // Count will store the number of // children of the current node int count = 0; for (int i = 0; i < root.child.Count; i++) { // If any child of the current node doesn't // satify the condition of being // a binary tree node if (!isBinaryTree(root.child[i])) return false; // Increment the count of children count++; // If current node has more // than 2 children if (count > 2) return false; } return true;} // Driver codepublic static void Main(String[] args) { Node root = newNode('A'); (root.child).Add(newNode('B')); (root.child).Add(newNode('C')); (root.child[0].child).Add(newNode('D')); (root.child[0].child).Add(newNode('E')); (root.child[1].child).Add(newNode('F')); if (isBinaryTree(root)) Console.WriteLine("Yes"); else Console.WriteLine("No");}} // This code is contributed by 29AjayKumar |
Output:
Yes
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