# Count nodes with two children at level L in a Binary Tree

Given a Binary tree, the task is to count the number of nodes with two children at a given level L.

Examples:

```Input:
1
/  \
2    3
/ \    \
4   5    6
/    / \
7    8   9
L = 2
Output: 1

Input:
20
/   \
8     22
/ \    / \
5   3  4   25
/ \  / \     \
1  10 2  14    6
L = 3
Output: 2
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Initialize a variable count = 0. Recursively traverse the tree in a level order manner. If the current level is same as the given level, then check whether the current node has two children. If it has two children then increment the variable count.

Below is the implementation of the above approach:

## C++

 `// C++ program to find number of full nodes ` `// at a given level ` `#include ` `using` `namespace` `std; ` ` `  `// A binary tree node ` `struct` `Node { ` `    ``int` `data; ` `    ``struct` `Node *left, *right; ` `}; ` ` `  `// Utility function to allocate memory for a new node ` `struct` `Node* newNode(``int` `data) ` `{ ` `    ``struct` `Node* node = ``new` `(``struct` `Node); ` `    ``node->data = data; ` `    ``node->left = node->right = NULL; ` `    ``return` `(node); ` `} ` ` `  `// Function that returns the height of binary tree ` `int` `height(``struct` `Node* root) ` `{ ` `    ``if` `(root == NULL) ` `        ``return` `0; ` ` `  `    ``int` `lheight = height(root->left); ` `    ``int` `rheight = height(root->right); ` ` `  `    ``return` `max(lheight, rheight) + 1; ` `} ` ` `  `// Level Order traversal to find the number of nodes ` `// having two children ` `void` `LevelOrder(``struct` `Node* root, ``int` `level, ``int``& count) ` `{ ` `    ``if` `(root == NULL) ` `        ``return``; ` ` `  `    ``if` `(level == 1 && root->left && root->right) ` `        ``count++; ` ` `  `    ``else` `if` `(level > 1) { ` `        ``LevelOrder(root->left, level - 1, count); ` `        ``LevelOrder(root->right, level - 1, count); ` `    ``} ` `} ` ` `  `// Returns the number of full nodes ` `// at a given level ` `int` `CountFullNodes(``struct` `Node* root, ``int` `L) ` `{ ` `    ``// Stores height of tree ` `    ``int` `h = height(root); ` ` `  `    ``// Stores count of nodes at a given level ` `    ``// that have two children ` `    ``int` `count = 0; ` ` `  `    ``LevelOrder(root, L, count); ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``struct` `Node* root = newNode(7); ` `    ``root->left = newNode(5); ` `    ``root->right = newNode(6); ` `    ``root->left->left = newNode(8); ` `    ``root->left->right = newNode(1); ` `    ``root->left->left->left = newNode(2); ` `    ``root->left->left->right = newNode(11); ` `    ``root->right->left = newNode(3); ` `    ``root->right->right = newNode(9); ` `    ``root->right->right->right = newNode(13); ` `    ``root->right->right->left = newNode(10); ` `    ``root->right->right->right->left = newNode(4); ` `    ``root->right->right->right->right = newNode(12); ` ` `  `    ``int` `L = 3; ` ` `  `    ``cout << CountFullNodes(root, L); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find number of full nodes  ` `// at a given level  ` `class` `GFG ` `{ ` ` `  `//INT class ` `static` `class` `INT ` `{ ` `    ``int` `a; ` `} ` ` `  `// A binary tree node  ` `static` `class` `Node  ` `{  ` `    ``int` `data;  ` `    ``Node left, right;  ` `};  ` ` `  `// Utility function to allocate memory for a new node  ` `static` `Node newNode(``int` `data)  ` `{  ` `    ``Node node = ``new` `Node();  ` `    ``node.data = data;  ` `    ``node.left = node.right = ``null``;  ` `    ``return` `(node);  ` `}  ` ` `  `// Function that returns the height of binary tree  ` `static` `int` `height(Node root)  ` `{  ` `    ``if` `(root == ``null``)  ` `        ``return` `0``;  ` ` `  `    ``int` `lheight = height(root.left);  ` `    ``int` `rheight = height(root.right);  ` ` `  `    ``return` `Math.max(lheight, rheight) + ``1``;  ` `}  ` ` `  `// Level Order traversal to find the number of nodes  ` `// having two children  ` `static` `void` `LevelOrder( Node root, ``int` `level, INT count)  ` `{  ` `    ``if` `(root == ``null``)  ` `        ``return``;  ` ` `  `    ``if` `(level == ``1` `&& root.left!=``null` `&& root.right!=``null``)  ` `        ``count.a++;  ` ` `  `    ``else` `if` `(level > ``1``) ` `    ``{  ` `        ``LevelOrder(root.left, level - ``1``, count);  ` `        ``LevelOrder(root.right, level - ``1``, count);  ` `    ``}  ` `}  ` ` `  `// Returns the number of full nodes  ` `// at a given level  ` `static` `int` `CountFullNodes( Node root, ``int` `L)  ` `{  ` `    ``// Stores height of tree  ` `    ``int` `h = height(root);  ` ` `  `    ``// Stores count of nodes at a given level  ` `    ``// that have two children  ` `    ``INT count =``new` `INT(); ` `    ``count.a = ``0``;  ` ` `  `    ``LevelOrder(root, L, count);  ` ` `  `    ``return` `count.a;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String args[]) ` `{  ` `    ``Node root = newNode(``7``);  ` `    ``root.left = newNode(``5``);  ` `    ``root.right = newNode(``6``);  ` `    ``root.left.left = newNode(``8``);  ` `    ``root.left.right = newNode(``1``);  ` `    ``root.left.left.left = newNode(``2``);  ` `    ``root.left.left.right = newNode(``11``);  ` `    ``root.right.left = newNode(``3``);  ` `    ``root.right.right = newNode(``9``);  ` `    ``root.right.right.right = newNode(``13``);  ` `    ``root.right.right.left = newNode(``10``);  ` `    ``root.right.right.right.left = newNode(``4``);  ` `    ``root.right.right.right.right = newNode(``12``);  ` ` `  `    ``int` `L = ``3``;  ` ` `  `    ``System.out.print( CountFullNodes(root, L));  ` ` `  `}  ` `} ` ` `  `// This code is contributed by Arnab Kundu `

## C#

 `// C# program to find number of full nodes  ` `// at a given level ` `using` `System; ` ` `  `class` `GFG  ` `{  ` ` `  `// INT class  ` `public` `class` `INT  ` `{  ` `    ``public` `int` `a;  ` `}  ` ` `  `// A binary tree node  ` `public` `class` `Node  ` `{  ` `    ``public` `int` `data;  ` `    ``public` `Node left, right;  ` `};  ` ` `  `// Utility function to allocate memory for a new node  ` `static` `Node newNode(``int` `data)  ` `{  ` `    ``Node node = ``new` `Node();  ` `    ``node.data = data;  ` `    ``node.left = node.right = ``null``;  ` `    ``return` `(node);  ` `}  ` ` `  `// Function that returns the height of binary tree  ` `static` `int` `height(Node root)  ` `{  ` `    ``if` `(root == ``null``)  ` `        ``return` `0;  ` ` `  `    ``int` `lheight = height(root.left);  ` `    ``int` `rheight = height(root.right);  ` ` `  `    ``return` `Math.Max(lheight, rheight) + 1;  ` `}  ` ` `  `// Level Order traversal to find the number of nodes  ` `// having two children  ` `static` `void` `LevelOrder( Node root, ``int` `level, INT count)  ` `{  ` `    ``if` `(root == ``null``)  ` `        ``return``;  ` ` `  `    ``if` `(level == 1 && root.left!=``null` `&& root.right!=``null``)  ` `        ``count.a++;  ` ` `  `    ``else` `if` `(level > 1)  ` `    ``{  ` `        ``LevelOrder(root.left, level - 1, count);  ` `        ``LevelOrder(root.right, level - 1, count);  ` `    ``}  ` `}  ` ` `  `// Returns the number of full nodes  ` `// at a given level  ` `static` `int` `CountFullNodes( Node root, ``int` `L)  ` `{  ` `    ``// Stores height of tree  ` `    ``int` `h = height(root);  ` ` `  `    ``// Stores count of nodes at a given level  ` `    ``// that have two children  ` `    ``INT count =``new` `INT();  ` `    ``count.a = 0;  ` ` `  `    ``LevelOrder(root, L, count);  ` ` `  `    ``return` `count.a;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main(String []args)  ` `{  ` `    ``Node root = newNode(7);  ` `    ``root.left = newNode(5);  ` `    ``root.right = newNode(6);  ` `    ``root.left.left = newNode(8);  ` `    ``root.left.right = newNode(1);  ` `    ``root.left.left.left = newNode(2);  ` `    ``root.left.left.right = newNode(11);  ` `    ``root.right.left = newNode(3);  ` `    ``root.right.right = newNode(9);  ` `    ``root.right.right.right = newNode(13);  ` `    ``root.right.right.left = newNode(10);  ` `    ``root.right.right.right.left = newNode(4);  ` `    ``root.right.right.right.right = newNode(12);  ` ` `  `    ``int` `L = 3;  ` ` `  `    ``Console.Write( CountFullNodes(root, L));  ` ` `  `}  ` `}  ` ` `  `// This code has been contributed by 29AjayKumar `

Output:

```2
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : andrew1234, 29AjayKumar