# 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. ## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 ` `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; ` `} `

## 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 `

## 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) `

## 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 of above code is O(n) as the code does a simple tree traversal.

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Improved By : shrikanth13

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