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# Print all full nodes in a Binary Tree

• Difficulty Level : Easy
• Last Updated : 08 Jul, 2022

Given a binary tree, print all nodes will are full nodes. Full Nodes are nodes which has both left and right children as non-empty.

Examples:

```Input :    10
/  \
8    2
/ \   /
3   5 7
Output : 10 8

Input :   1
/ \
2   3
/ \
4   6
Output : 1 3```

This is a simple problem. We do any of the traÂ­verÂ­sals (Inorder, PreÂ­order, PosÂ­torder, level order traversal) and keep printing nodes that have mode left and right children as non-NULL.

Implementation:

## C++

 `// A C++ program to find the all full nodes in` `// a given binary tree` `#include ` `using` `namespace` `std;`   `struct` `Node` `{` `    ``int` `data;` `    ``struct` `Node *left, *right;` `};`   `Node *newNode(``int` `data)` `{` `    ``Node *temp = ``new` `Node;` `    ``temp->data = data;` `    ``temp->left = temp->right = NULL;` `    ``return` `temp;` `}`   `// Traverses given tree in Inorder fashion and` `// prints all nodes that have both children as` `// non-empty.` `void` `findFullNode(Node *root)` `{` `    ``if` `(root != NULL)` `    ``{` `        ``findFullNode(root->left);` `        ``if` `(root->left != NULL && root->right != NULL)` `            ``cout << root->data << ``" "``;` `        ``findFullNode(root->right);` `    ``}` `}`   `// Driver program to test above function` `int` `main()` `{` `    ``Node* root = newNode(1);` `    ``root->left = newNode(2);` `    ``root->right = newNode(3);` `    ``root->left->left = newNode(4);` `    ``root->right->left = newNode(5);` `    ``root->right->right = newNode(6);` `    ``root->right->left->right = newNode(7);` `    ``root->right->right->right = newNode(8);` `    ``root->right->left->right->left = newNode(9);` `    ``findFullNode(root);` `    ``return` `0;` `}`

## Java

 `// Java program to find the all full nodes in ` `// a given binary tree ` `public` `class` `FullNodes {`   `    ``// Traverses given tree in Inorder fashion and ` `    ``// prints all nodes that have both children as ` `    ``// non-empty. ` `    ``public` `static` `void` `findFullNode(Node root) ` `    ``{ ` `        ``if` `(root != ``null``) ` `        ``{ ` `            ``findFullNode(root.left); ` `            ``if` `(root.left != ``null` `&& root.right != ``null``) ` `                ``System.out.print(root.data+``" "``);` `            ``findFullNode(root.right); ` `        ``} ` `    ``} `   `    ``public` `static` `void` `main(String args[]) {` `        ``Node root = ``new` `Node(``1``); ` `        ``root.left = ``new` `Node(``2``); ` `        ``root.right = ``new` `Node(``3``); ` `        ``root.left.left = ``new` `Node(``4``); ` `        ``root.right.left = ``new` `Node(``5``); ` `        ``root.right.right = ``new` `Node(``6``); ` `        ``root.right.left.right = ``new` `Node(``7``); ` `        ``root.right.right.right = ``new` `Node(``8``); ` `        ``root.right.left.right.left = ``new` `Node(``9``); ` `        ``findFullNode(root); ` `    ``}` `}`   `/* A binary tree node */` `class` `Node ` `{ ` `    ``int` `data; ` `    ``Node left, right; ` `    ``Node(``int` `data)` `    ``{` `        ``left=right=``null``;` `        ``this``.data=data;` `    ``}` `}; ` `//This code is contributed by Gaurav Tiwari`

## Python3

 `# Python3 program to find the all ` `# full nodes in a given binary tree`   `# Binary Tree Node ` `""" utility that allocates a newNode ` `with the given key """` `class` `newNode: `   `    ``# Construct to create a newNode ` `    ``def` `__init__(``self``, key): ` `        ``self``.data ``=` `key` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None`   `# Traverses given tree in Inorder ` `# fashion and prints all nodes that ` `# have both children as non-empty. ` `def` `findFullNode(root) :`   `    ``if` `(root !``=` `None``) :` `    `  `        ``findFullNode(root.left) ` `        ``if` `(root.left !``=` `None` `and` `            ``root.right !``=` `None``) :` `            ``print``(root.data, end ``=` `" "``) ` `        ``findFullNode(root.right) `   `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``:`   `    ``root ``=` `newNode(``1``) ` `    ``root.left ``=` `newNode(``2``) ` `    ``root.right ``=` `newNode(``3``) ` `    ``root.left.left ``=` `newNode(``4``) ` `    ``root.right.left ``=` `newNode(``5``) ` `    ``root.right.right ``=` `newNode(``6``) ` `    ``root.right.left.right ``=` `newNode(``7``) ` `    ``root.right.right.right ``=` `newNode(``8``) ` `    ``root.right.left.right.left ``=` `newNode(``9``) ` `    ``findFullNode(root)`   `# This code is contributed by` `# Shubham Singh(SHUBHAMSINGH10)`

## C#

 `// C# program to find the all full nodes in ` `// a given binary tree ` `using` `System;`   `public` `class` `FullNodes` `{`   `    ``// Traverses given tree in Inorder fashion and ` `    ``// prints all nodes that have both children as ` `    ``// non-empty. ` `    ``static` `void` `findFullNode(Node root) ` `    ``{ ` `        ``if` `(root != ``null``) ` `        ``{ ` `            ``findFullNode(root.left); ` `            ``if` `(root.left != ``null` `&& root.right != ``null``) ` `                ``Console.Write(root.data + ``" "``);` `            ``findFullNode(root.right); ` `        ``} ` `    ``} `   `    ``public` `static` `void` `Main(String []args) ` `    ``{` `        ``Node root = ``new` `Node(1); ` `        ``root.left = ``new` `Node(2); ` `        ``root.right = ``new` `Node(3); ` `        ``root.left.left = ``new` `Node(4); ` `        ``root.right.left = ``new` `Node(5); ` `        ``root.right.right = ``new` `Node(6); ` `        ``root.right.left.right = ``new` `Node(7); ` `        ``root.right.right.right = ``new` `Node(8); ` `        ``root.right.left.right.left = ``new` `Node(9); ` `        ``findFullNode(root); ` `    ``}` `}`   `/* A binary tree node */` `class` `Node ` `{ ` `    ``public` `int` `data; ` `    ``public` `Node left, right; ` `    ``public` `Node(``int` `data)` `    ``{` `        ``left = right = ``null``;` `        ``this``.data = data;` `    ``}` `};`   `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output

`1 3 `

Time Complexity : O(n)
Space complexity: O(n) for Recursive Stack Space in case of Skewed Tree

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