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

• Difficulty Level : Easy
• Last Updated : 06 Aug, 2021

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```

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`

## Python3

 `# Python3 program to find number of``# full nodes at a given level` `# INT class``class` `INT``:`` ` `    ``def` `__init__(``self``):``        ` `        ``self``.a ``=` `0` `# A binary tree node``class` `Node:``    ` `    ``def` `__init__(``self``, data):``        ` `        ``self``.left ``=` `None``        ``self``.right ``=` `None``        ``self``.data ``=` `data`` ` `# Utility function to allocate``# memory for a new node``def` `newNode(data):` `    ``node ``=` `Node(data)``    ` `    ``return` `node` `# Function that returns the``# height of binary tree``def` `height(root):` `    ``if` `(root ``=``=` `None``):``        ``return` `0``;`` ` `    ``lheight ``=` `height(root.left);``    ``rheight ``=` `height(root.right);`` ` `    ``return` `max``(lheight, rheight) ``+` `1``;` `# Level Order traversal to find the``# number of nodes having two children``def` `LevelOrder(root, level, count):` `    ``if` `(root ``=``=` `None``):``        ``return``;`` ` `    ``if` `(level ``=``=` `1` `and``        ``root.left !``=` `None` `and``       ``root.right !``=` `None``):``        ``count.a ``+``=` `1`` ` `    ``elif` `(level > ``1``):``        ``LevelOrder(root.left,``                   ``level ``-` `1``, count);``        ``LevelOrder(root.right,``                   ``level ``-` `1``, count);`` ` `# Returns the number of full nodes``# at a given level``def` `CountFullNodes(root, L):` `    ``# Stores height of tree``    ``h ``=` `height(root);`` ` `    ``# Stores count of nodes at a``    ``# given level that have two children``    ``count ``=` `INT``()`` ` `    ``LevelOrder(root, L, count);`` ` `    ``return` `count.a` `# Driver code   ``if` `__name__``=``=``"__main__"``:``    ` `    ``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``);`` ` `    ``L ``=` `3``;`` ` `    ``print``(CountFullNodes(root, L))``    ` `# This code is contributed by rutvik_56`

## 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 is contributed by 29AjayKumar`

## Javascript

 ``
Output:

`2`

Time Complexity: O(N)
Auxiliary Space: O(N)

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