# Program to count leaf nodes in a binary tree

• Difficulty Level : Easy
• Last Updated : 30 Jan, 2023

A node is a leaf node if both left and right child nodes of it are NULL.
Here is an algorithm to get the leaf node count.

```getLeafCount(node)
1) If node is NULL then return 0.
2) Else If left and right child nodes are NULL return 1.
3) Else recursively calculate leaf count of the tree using below formula.
Leaf count of a tree = Leaf count of left subtree +
Leaf count of right subtree```

Leaf count for the above tree is 3.

Implementation:

## C++

 `// C++ implementation to find leaf``// count of a given Binary tree``#include ``using` `namespace` `std;` `/* A binary tree node has data,``pointer to left child and``a pointer to right child */``struct` `node``{``    ``int` `data;``    ``struct` `node* left;``    ``struct` `node* right;``};` `/* Function to get the count``of leaf nodes in a binary tree*/``unsigned ``int` `getLeafCount(``struct` `node* node)``{``    ``if``(node == NULL)    ``        ``return` `0;``    ``if``(node->left == NULL && node->right == NULL)``        ``return` `1;        ``    ``else``        ``return` `getLeafCount(node->left)+``            ``getLeafCount(node->right);``}` `/* Helper function that allocates a new node with the``given data and NULL left and right pointers. */``struct` `node* newNode(``int` `data)``{``    ``struct` `node* node = (``struct` `node*)``                    ``malloc``(``sizeof``(``struct` `node));``    ``node->data = data;``    ``node->left = NULL;``    ``node->right = NULL;``    ` `return``(node);``}` `/*Driver code*/``int` `main()``{``    ``/*create a tree*/``    ``struct` `node *root = newNode(1);``    ``root->left = newNode(2);``    ``root->right = newNode(3);``    ``root->left->left = newNode(4);``    ``root->left->right = newNode(5);``    ` `/*get leaf count of the above created tree*/``cout << ``"Leaf count of the tree is : "``<<``                ``getLeafCount(root) << endl;``return` `0;``}` `// This code is contributed by SHUBHAMSINGH10`

## C

 `// C implementation to find leaf count of a given Binary tree``#include ``#include ` `/* A binary tree node has data, pointer to left child``   ``and a pointer to right child */``struct` `node``{``    ``int` `data;``    ``struct` `node* left;``    ``struct` `node* right;``};` `/* Function to get the count of leaf nodes in a binary tree*/``unsigned ``int` `getLeafCount(``struct` `node* node)``{``  ``if``(node == NULL)      ``    ``return` `0;``  ``if``(node->left == NULL && node->right==NULL)     ``    ``return` `1;           ``  ``else``    ``return` `getLeafCount(node->left)+``           ``getLeafCount(node->right);     ``}` `/* Helper function that allocates a new node with the``   ``given data and NULL left and right pointers. */``struct` `node* newNode(``int` `data)``{``  ``struct` `node* node = (``struct` `node*)``                       ``malloc``(``sizeof``(``struct` `node));``  ``node->data = data;``  ``node->left = NULL;``  ``node->right = NULL;``  ` `  ``return``(node);``}` `/*Driver program to test above functions*/` `int` `main()``{``  ``/*create a tree*/` `  ``struct` `node *root = newNode(1);``  ``root->left        = newNode(2);``  ``root->right       = newNode(3);``  ``root->left->left  = newNode(4);``  ``root->left->right = newNode(5);   ``  ` `  ``/*get leaf count of the above created tree*/``  ``printf``(``"Leaf count of the tree is %d"``, getLeafCount(root));``  ` `  ``getchar``();``  ``return` `0;``}`

## Java

 `// Java implementation to find leaf count of a given Binary tree`` ` `/* Class containing left and right child of current`` ``node and key value*/``class` `Node``{``    ``int` `data;``    ``Node left, right;`` ` `    ``public` `Node(``int` `item)``    ``{``        ``data = item;``        ``left = right = ``null``;``    ``}``}`` ` `public` `class` `BinaryTree``{``    ``//Root of the Binary Tree``    ``Node root;``     ` `    ``/* Function to get the count of leaf nodes in a binary tree*/``    ``int` `getLeafCount()``    ``{``        ``return` `getLeafCount(root);``    ``}`` ` `    ``int` `getLeafCount(Node node)``    ``{``        ``if` `(node == ``null``)``            ``return` `0``;``        ``if` `(node.left == ``null` `&& node.right == ``null``)``            ``return` `1``;``        ``else``            ``return` `getLeafCount(node.left) + getLeafCount(node.right);``    ``}`` ` `    ``/* Driver program to test above functions */``    ``public` `static` `void` `main(String args[])``    ``{``        ``/* create a tree */``        ``BinaryTree tree = ``new` `BinaryTree();``        ``tree.root = ``new` `Node(``1``);``        ``tree.root.left = ``new` `Node(``2``);``        ``tree.root.right = ``new` `Node(``3``);``        ``tree.root.left.left = ``new` `Node(``4``);``        ``tree.root.left.right = ``new` `Node(``5``);``         ` `        ``/* get leaf count of the above tree */``        ``System.out.println(``"The leaf count of binary tree is : "``                             ``+ tree.getLeafCount());``    ``}``}` `// This code has been contributed by Mayank Jaiswal(mayank_24)`

## Python3

 `# Python program to count leaf nodes in Binary Tree` `# A Binary tree node``class` `Node:``    ` `    ``# Constructor to create a new node``    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `# Function to get the count of leaf nodes in binary tree``def` `getLeafCount(node):``    ``if` `node ``is` `None``:``        ``return` `0``    ``if``(node.left ``is` `None` `and` `node.right ``is` `None``):``        ``return` `1``    ``else``:``        ``return` `getLeafCount(node.left) ``+` `getLeafCount(node.right)`  `# Driver program to test above function``root ``=` `Node(``1``)``root.left ``=` `Node(``2``)``root.right ``=` `Node(``3``)``root.left.left ``=` `Node(``4``)``root.left.right ``=` `Node(``5``)` `print` `(``"Leaf count of the tree is %d"` `%``(getLeafCount(root)))` `#This code is contributed by Nikhil Kumar Singh(nickzuck_007)`

## C#

 `using` `System;` `// C# implementation to find leaf count of a given Binary tree` `/* Class containing left and right child of current `` ``node and key value*/``public` `class` `Node``{``    ``public` `int` `data;``    ``public` `Node left, right;` `    ``public` `Node(``int` `item)``    ``{``        ``data = item;``        ``left = right = ``null``;``    ``}``}` `public` `class` `BinaryTree``{``    ``//Root of the Binary Tree``    ``public` `Node root;` `    ``/* Function to get the count of leaf nodes in a binary tree*/``    ``public` `virtual` `int` `LeafCount``    ``{``        ``get``        ``{``            ``return` `getLeafCount(root);``        ``}``    ``}` `    ``public` `virtual` `int` `getLeafCount(Node node)``    ``{``        ``if` `(node == ``null``)``        ``{``            ``return` `0;``        ``}``        ``if` `(node.left == ``null` `&& node.right == ``null``)``        ``{``            ``return` `1;``        ``}``        ``else``        ``{``            ``return` `getLeafCount(node.left) + getLeafCount(node.right);``        ``}``    ``}` `    ``/* Driver program to test above functions */``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``/* create a tree */``        ``BinaryTree tree = ``new` `BinaryTree();``        ``tree.root = ``new` `Node(1);``        ``tree.root.left = ``new` `Node(2);``        ``tree.root.right = ``new` `Node(3);``        ``tree.root.left.left = ``new` `Node(4);``        ``tree.root.left.right = ``new` `Node(5);` `        ``/* get leaf count of the above tree */``        ``Console.WriteLine(``"The leaf count of binary tree is : "` `+ tree.LeafCount);``    ``}``}` `  ``// This code is contributed by Shrikant13`

## Javascript

 ``

Output

```Leaf count of the tree is : 3
```

Time & Space Complexities: Since this program is similar to traversal of tree, time and space complexities will be same as Tree traversal (Please see our Tree Traversal post for details)

Another Approach(Iterative):
The given problem can be solved by using the Level Order Traversal(https://www.geeksforgeeks.org/level-order-tree-traversal/). Follow the steps below to solve the problem:

1) Create a queue(q) and initialize count variable with 0, and store the nodes in q along wise level order and iterate for next level.
2) Perform level order traversal and check if current node is a leaf node(don’t have right and left child) then increment the count variable.
3) After completing the above steps, return count variable.

Below is the implementation of above approach:

## C++

 `// C++ implementation to find leaf``// count of a given Binary tree``#include ``using` `namespace` `std;` `// A binary tree node``struct` `Node{``    ``int` `data;``    ``struct` `Node* left;``    ``struct` `Node* right;``};` `// utility funciton to get the new node``Node* newNode(``int` `data){``    ``Node *new_node = ``new` `Node();``    ``new_node->data = data;``    ``new_node->left = NULL;``    ``new_node->right = NULL;``    ``return` `new_node;``}` `// function to get count of leaf nodes``int` `getLeafCount(Node* root){``    ``// initializing queue for level order traversal``    ``queue q;``    ``q.push(root);``    ``// initializing count variable``    ``int` `count = 0;``    ``while``(!q.empty()){``        ``Node* temp = q.front();``        ``q.pop();``        ``if``(temp->left == NULL && temp->right == NULL)``            ``count++;``        ``if``(temp->left) q.push(temp->left);``        ``if``(temp->right) q.push(temp->right);``    ``}``    ``return` `count;``}`  `// driver code to test above function``int` `main(){``    ``struct` `Node *root = newNode(1);``    ``root->left = newNode(2);``    ``root->right = newNode(3);``    ``root->left->left = newNode(4);``    ``root->left->right = newNode(5);` `    ``// function call``    ``cout << ``"Leaf count of the tree is : "``<< getLeafCount(root) << endl;``    ``return` `0;``}` `// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)`

Output

```Leaf count of the tree is : 3
```

Time Complexity: O(N) where N is the number of nodes in binary tree.
Auxiliary Space: O(N) due to queue data structure.

Please write comments if you find any bug in the above programs/algorithms or other ways to solve the same problem.

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