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 root and check if the node has one child, if yes then print the only child of that node. If node has both children, then recur for both the children.
C++
/* Program to find singles in a given binary tree */#include <bits/stdc++.h> using namespace std; // A Binary Tree Node struct node { struct node *left, *right; int key; }; // Utility function to create a new tree node node* 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 sibling void 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 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); printSingles(root); return 0; } |
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Java
// Java program to print all nodes that don't have sibling // A binary tree node class 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 |
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Python
# Python3 program to find singles in a given binary tree # A Binary Tree Node class 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 sibling def 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 function root = 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) |
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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 |
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Output:
4 5 6
Time Complexity of above code is O(n) as the code does a simple tree traversal.
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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Improved By : shrikanth13
