Print the nodes having exactly one child in a Binary tree
Given a binary tree, the task is to print all the nodes having exactly one child. Print “-1” if no such node exists.
Examples:
Input:
2
/ \
3 5
/ / \
7 8 6
Output: 3
Explanation:
There is only one node having
single child that is 3.
Input:
9
/ \
7 8
/ \
4 3
Output: -1
Explanation:
There is no node having exactly one
child in the binary tree.
Approach: The idea is to traverse the tree in the inorder traversal and at each step of the traversal check that if the node is having exactly one child. Then append that node into a result array to keep track of such nodes. After the traversal, simply print each element of this result array.
Below is the implementation of the above approach:
C++
// C++ implementation to print// the nodes having a single child#include <bits/stdc++.h>using namespace std;// Class of the Binary Tree nodestruct Node{ int data; Node *left, *right; Node(int x) { data = x; left = right = NULL; }};vector<int> lst;// Function to find the nodes// having single childvoid printNodesOneChild(Node* root){ // Base case if (root == NULL) return; // Condition to check if the // node is having only one child if (root->left != NULL && root->right == NULL || root->left == NULL && root->right != NULL) { lst.push_back(root->data); } // Traversing the left child printNodesOneChild(root -> left); // Traversing the right child printNodesOneChild(root -> right);}//Driver codeint main(){ // Constructing the binary tree Node *root = new Node(2); root -> left = new Node(3); root -> right = new Node(5); root -> left -> left = new Node(7); root -> right -> left = new Node(8); root -> right -> right = new Node(6); // Function calling printNodesOneChild(root); // Condition to check if there is // no such node having single child if (lst.size() == 0) printf("-1"); else { for(int value : lst) { cout << (value) << endl; } }}// This code is contributed by Mohit Kumar 29 |
Java
// Java implementation to print// the nodes having a single childimport java.util.Vector;// Class of the Binary Tree nodeclass Node{ int data; Node left, right; Node(int data) { this.data = data; }}class GFG{// List to keep track of nodes// having single childstatic Vector<Integer> list = new Vector<>();// Function to find the nodes // having single child static void printNodesOneChild(Node root){ // Base case if (root == null) return; // Condition to check if the node // is having only one child if (root.left != null && root.right == null || root.left == null && root.right != null) { list.add(root.data); } // Traversing the left child printNodesOneChild(root.left); // Traversing the right child printNodesOneChild(root.right);}// Driver codepublic static void main(String[] args){ // Constructing the binary tree Node root = new Node(2); root.left = new Node(3); root.right = new Node(5); root.left.left = new Node(7); root.right.left = new Node(8); root.right.right = new Node(6); // Function calling printNodesOneChild(root); // Condition to check if there is // no such node having single child if (list.size() == 0) System.out.println(-1); else { for(int value : list) { System.out.println(value); } }}}// This code is contributed by sumit_9 |
Python3
# Python3 implementation to print# the nodes having a single child# Class of the Binary Tree nodeclass node: # Constructor to construct # the node of the Binary tree def __init__(self, val): self.val = val self.left = None self.right = None # List to keep track of# nodes having single childsingle_child_nodes = []# Function to find the nodes# having single childdef printNodesOneChild(root): # Base Case if not root: return # Condition to check if the node # is having only one child if not root.left and root.right: single_child_nodes.append(root) elif root.left and not root.right: single_child_nodes.append(root) # Traversing the left child printNodesOneChild(root.left) # Traversing the right child printNodesOneChild(root.right) return# Driver Codeif __name__ == "__main__": # Condtruction of Binary Tree root = node(2) root.left = node(3) root.right = node(5) root.left.left = node(7) root.right.left = node(8) root.right.right = node(6) # Function Call printNodesOneChild(root) # Condition to check if there is # no such node having single child if not len(single_child_nodes): print(-1) else: for i in single_child_nodes: print(i.val, end = " ") print() |
C#
// C# implementation to print// the nodes having a single childusing System;using System.Collections;class GFG{ // Structure of binary treepublic class Node { public Node left; public Node right; public int data;};// Function to create a new nodestatic Node newNode(int key){ Node node = new Node(); node.left = node.right = null; node.data = key; return node;}// List to keep track of nodes// having single childstatic ArrayList list = new ArrayList(); // Function to find the nodes // having single child static void printNodesOneChild(Node root){ // Base case if (root == null) return; // Condition to check if the node // is having only one child if (root.left != null && root.right == null || root.left == null && root.right != null) { list.Add(root.data); } // Traversing the left child printNodesOneChild(root.left); // Traversing the right child printNodesOneChild(root.right);}// Driver codestatic public void Main(){ // Constructing the binary tree Node root = newNode(2); root.left = newNode(3); root.right = newNode(5); root.left.left = newNode(7); root.right.left = newNode(8); root.right.right = newNode(6); // Function calling printNodesOneChild(root); // Condition to check if there is // no such node having single child if (list.Count == 0) Console.WriteLine(-1); else { foreach(int value in list) { Console.WriteLine(value); } }}}// This code is contributed by offbeat |
Output
3
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