Print all nodes that don’t have sibling
Given a Binary Tree, print all nodes that don’t have a sibling (a sibling is a node that has same parent. In a Binary Tree, there can be at most one sibling). Root should not be printed as root cannot have a sibling.
For example, the output should be “4 5 6” for the following tree.
This is a typical tree traversal question. We start from the root and check if the node has one child, if yes then print the only child of that node. If the node has both children, then recur for both the children.
Below is the implementation of above approach:
C++
/* Program to find singles in a given binary tree */#include <bits/stdc++.h>using namespace std;// A Binary Tree Nodestruct node{ struct node *left, *right; int key;};// Utility function to create a new tree nodenode* newNode(int key){ node *temp = new node; temp->key = key; temp->left = temp->right = NULL; return temp;}// Function to print all non-root nodes // that don't have a siblingvoid printSingles(struct node *root){ // Base case if (root == NULL) return; // If this is an internal node, recur for left // and right subtrees if (root->left != NULL && root->right != NULL) { printSingles(root->left); printSingles(root->right); } // If left child is NULL and right is not, // print right child // and recur for right child else if (root->right != NULL) { cout << root->right->key << " "; printSingles(root->right); } // If right child is NULL and left is // not, print left child // and recur for left child else if (root->left != NULL) { cout << root->left->key << " "; printSingles(root->left); }}// Driver program to test above functionsint main(){ // Let us create binary tree // given in the above example node *root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->right = newNode(4); root->right->left = newNode(5); root->right->left->left = newNode(6); printSingles(root); return 0;} |
Java
// Java program to print all nodes// that don't have sibling// A binary tree nodeclass Node { int data; Node left, right; Node(int item) { data = item; left = right = null; }}class BinaryTree { Node root; // Function to print all non-root nodes // that don't have a sibling void printSingles(Node node) { // Base case if (node == null) return; // If this is an internal node, recur for left // and right subtrees if (node.left != null && node.right != null) { printSingles(node.left); printSingles(node.right); } // If left child is NULL and right // is not, print right child // and recur for right child else if (node.right != null) { System.out.print(node.right.data + " "); printSingles(node.right); } // If right child is NULL and left // is not, print left child // and recur for left child else if (node.left != null) { System.out.print( node.left.data + " "); printSingles(node.left); }} // Driver program to test the above functions public static void main(String args[]) { BinaryTree tree = new BinaryTree(); /* Let us construct the tree shown in above diagram */ tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.right = new Node(4); tree.root.right.left = new Node(5); tree.root.right.left.right = new Node(6); tree.printSingles(tree.root); }}// This code has been contributed by Mayank Jaiswal |
Python
# Python3 program to find singles in a given binary tree# A Binary Tree Nodeclass Node: # A constructor to create new tree node def __init__(self, key): self.key = key self.left = None self.right = None# Function to print all non-root nodes that don't have# a siblingdef printSingles(root): # Base Case if root is None: return # If this is an internal node , recur for left # and right subtrees if root.left is not None and root.right is not None: printSingles(root.left) printSingles(root.right) # If left child is NULL, and right is not, print # right child and recur for right child elif root.right is not None: print root.right.key, printSingles(root.right) # If right child is NULL and left is not, print # left child and recur for left child elif root.left is not None: print root.left.key, printSingles(root.left)# Driver program to test above functionroot = Node(1)root.left = Node(2)root.right = Node(3)root.left.right = Node(4)root.right.left = Node(5)root.right.left.left = Node(6)printSingles(root)# This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
using System;// C# program to print all nodes that don't have sibling // A binary tree node public class Node{ public int data; public Node left, right; public Node(int item) { data = item; left = right = null; }}public class BinaryTree{ public Node root; // Function to print all non-root nodes that don't have a sibling public virtual void printSingles(Node node) { // Base case if (node == null) { return; } // If this is an internal node, recur for left // and right subtrees if (node.left != null && node.right != null) { printSingles(node.left); printSingles(node.right); } // If left child is NULL and right is not, print right child // and recur for right child else if (node.right != null) { Console.Write(node.right.data + " "); printSingles(node.right); } // If right child is NULL and left is not, print left child // and recur for left child else if (node.left != null) { Console.Write(node.left.data + " "); printSingles(node.left); } } // Driver program to test the above functions public static void Main(string[] args) { BinaryTree tree = new BinaryTree(); /* Let us construct the tree shown in above diagram */ tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.right = new Node(4); tree.root.right.left = new Node(5); tree.root.right.left.right = new Node(6); tree.printSingles(tree.root); }}// This code is contributed by Shrikant13 |
Output:
4 5 6
Time Complexity: O(n)
Alternate implementation in iterative method:
We start from the root and check if the node has one child, if yes then print the only child of that node. If the node has both children, then push both the children in the queue.
Below is the implementation of the above approach:
C++14
// CPP program for above approach#include <bits/stdc++.h>using namespace std; // A Binary Tree Node struct node { struct node *left, *right; int data; }; // Utility function to // create a new tree node node* newNode(int key) { node *temp = new node; temp->data= key; temp->left = temp->right = NULL; return temp; } // Function to print all // non-root nodes that // don't have a sibling void printSingles(struct node *root) { // Base case if (root == NULL) return; queue<node *> q1; q1.push(root); int flag=0; vector<int> v; // While q1 is not empty while(q1.empty() == false) { struct node * temp=q1.front(); q1.pop(); // Check if temp->left is not // NULL and temp->right is NULL if(temp->left != NULL && temp->right == NULL) { flag=1; v.push_back(temp->left->data); } // Check if temp->left is equal // NULL and temp->right is not NULL if(temp->left == NULL && temp->right != NULL) { flag=1; v.push_back(temp->right->data); } // Check if temp->left is not // NULL if(temp->left != NULL) { q1.push(temp->left); } // Check if temp->right is not // NULL if(temp->right != NULL) { q1.push(temp->right); } } // Sort v in increasing order sort(v.begin(), v.end()); // Iterate i from 0 to v.size() - 1 for (int i = 0; i < v.size(); i++) { cout<< v[i] << " "; } // Check is v is empty if (v.size() == 0) { cout<<"-1"; } } // Driver program to test // above functions int main() { // Let us create binary tree // given in the above example node *root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->right = newNode(4); root->right->left = newNode(5); root->right->left->left = newNode(6); // Function Call printSingles(root); return 0; } |
Java
// JAVA program for above approachimport java.util.*;class GFG{ // A Binary Tree Node static class node { node left, right; int data; }; // Utility function to // create a new tree node static node newNode(int key) { node temp = new node(); temp.data = key; temp.left = temp.right = null; return temp; } // Function to print all // non-root nodes that // don't have a sibling static void printSingles(node root) { // Base case if (root == null) return; Queue<node > q1 = new LinkedList<>(); q1.add(root); int flag = 0; Vector<Integer> v = new Vector<>(); // While q1 is not empty while(q1.isEmpty() == false) { node temp = q1.peek(); q1.remove(); // Check if temp.left is not // null and temp.right is null if(temp.left != null && temp.right == null) { flag = 1; v.add(temp.left.data); } // Check if temp.left is equal // null and temp.right is not null if(temp.left == null && temp.right != null) { flag = 1; v.add(temp.right.data); } // Check if temp.left is not // null if(temp.left != null) { q1.add(temp.left); } // Check if temp.right is not // null if(temp.right != null) { q1.add(temp.right); } } // Sort v in increasing order Collections.sort(v); // Iterate i from 0 to v.size() - 1 for (int i = 0; i < v.size(); i++) { System.out.print( v.get(i) + " "); } // Check is v is empty if (v.size() == 0) { System.out.print("-1"); } } // Driver program to test // above functions public static void main(String[] args) { // Let us create binary tree // given in the above example node root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.right = newNode(4); root.right.left = newNode(5); root.right.left.left = newNode(6); // Function Call printSingles(root); }} // This code is contributed by aashish1995 |
C#
// C# program for above approachusing System;using System.Collections.Generic;class GFG{ // A Binary Tree Node public class node { public node left, right; public int data; }; // Utility function to // create a new tree node static node newNode(int key) { node temp = new node(); temp.data = key; temp.left = temp.right = null; return temp; } // Function to print all // non-root nodes that // don't have a sibling static void printSingles(node root) { // Base case if (root == null) return; Queue<node > q1 = new Queue<node>(); q1.Enqueue(root); int flag = 0; List<int> v = new List<int>(); // While q1 is not empty while(q1.Count != 0) { node temp = q1.Peek(); q1.Dequeue(); // Check if temp.left is not // null and temp.right is null if(temp.left != null && temp.right == null) { flag = 1; v.Add(temp.left.data); } // Check if temp.left is equal // null and temp.right is not null if(temp.left == null && temp.right != null) { flag = 1; v.Add(temp.right.data); } // Check if temp.left is not // null if(temp.left != null) { q1.Enqueue(temp.left); } // Check if temp.right is not // null if(temp.right != null) { q1.Enqueue(temp.right); } } // Sort v in increasing order v.Sort(); // Iterate i from 0 to v.Count - 1 for (int i = 0; i < v.Count; i++) { Console.Write( v[i] + " "); } // Check is v is empty if (v.Count == 0) { Console.Write("-1"); } } // Driver program to test // above functions public static void Main(String[] args) { // Let us create binary tree // given in the above example node root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.right = newNode(4); root.right.left = newNode(5); root.right.left.left = newNode(6); // Function Call printSingles(root); }} // This code is contributed by Rajput-Ji. |
Output
4 5 6
This article is compiled by Aman Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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