This is a C++ Program for counting the total number of leaf nodes present in a given Binary Search Tree.

We will be given a Binary Search Tree and we have to create a C++ program which counts the total number of leaf nodes present in it using recursion. A **leaf node** is one which has no child. It is also known as **internal node**.

**Case 1. Balanced Tree:** When the tree has same weight on both sides of root node.

25 / \ 19 29 / \ / \ 17 20 27 55

**Output: 4**

**Case 2. Right Skewed Tree:** When nodes at every level have just a right child.

1 \ 2 \ 3 \ 4 \ 5

**Output: 1**

**Case 3. Tree having just one node**

15

**Output: 1**

We can easily find the number of leaf nodes present in any tree using recursion. A leaf node is a node whose left and right child are NULL. We just need to check this single condition to determine whether the node is a leaf node or a non leaf (internal) node.

Here is source code of the C++ Program to count the total number of leaf nodes present in a given Binary Search Tree. The program is successfully compiled and tested using Codeblocks gnu/GCC compiler on windows 10. The program output is also shown below.

`/* C++ Program to find the number of leaf nodes in a Tree */`

`#include <iostream>`

using namespace std;

`struct node`

`{`

int info;

struct node *left, *right;

};

int count = 0;

`class BST`

`{`

public:

`/*`

`* Function to create new nodes`

`*/`

struct node *createnode(int key)

`{`

struct node *newnode = new node;

newnode->info = key;

newnode->left = NULL;

newnode->right = NULL;

return(newnode);

`}`

int leafnodes(struct node *newnode)

`{`

if(newnode != NULL)

`{`

leafnodes(newnode->left);

if((newnode->left == NULL) && (newnode->right == NULL))

`{`

count++;

`}`

leafnodes(newnode->right);

`}`

return count;

`}`

};

int main()

`{`

`/* Creating first Tree. */`

`BST t1,t2,t3;`

struct node *newnode = t1.createnode(25);

newnode->left = t1.createnode(19);

newnode->right = t1.createnode(29);

newnode->left->left = t1.createnode(17);

newnode->left->right = t1.createnode(20);

newnode->right->left = t1.createnode(27);

newnode->right->right = t1.createnode(55);

`/* Sample Tree 1. Balanced Tree`

`* 25`

`* / \`

`* 19 29`

`* / \ / \`

`* 17 20 27 55`

`*/`

cout<<"Number of leaf nodes in first Tree are\t"<<t1.leafnodes(newnode)<<endl;

count = 0;

`/* Creating second tree */`

struct node *node = t2.createnode(1);

node->right = t2.createnode(2);

node->right->right = t2.createnode(3);

node->right->right->right = t2.createnode(4);

node->right->right->right->right = t2.createnode(5);

`/* Sample Tree 2. Right Skewed Tree (Unbalanced).`

`* 1`

`* \`

`* 2`

`* \`

`* 3`

`* \`

`* 4`

`* \`

`* 5`

`*/`

cout<<"\nNumber of leaf nodes in second tree are\t"<<t2.leafnodes(node)<<endl;

count = 0;

`/*Creating third Tree. */`

struct node *root = t3.createnode(15);

`/* Sample Tree 3- Tree having just one root node.`

`* 15`

`*/`

cout<<"\nNumber of leaf nodes in third tree are\t"<<t3.leafnodes(root);

return 0;

`}`

**1.** In this program we have used recursion to find the total number of leaf nodes present in a tree.

**2.** A leaf Node is one whose left and right child are NULL. We have created a function called **leafnodes()** which takes in root of the tree as a parameter and returns the total number of leaf nodes it has.

**3.** The basic idea is to traverse the tree using any traversal so as to visit each and every node and check the condition for leaf node for each node, that is what we have done in leafnodes() function.

**4.** In the leafnodes() function we have used the inorder traversal, by first traversing the left subtree, then instead of printing the root->data as a second step of inorder traversal, we have checked the leaf node condition and then at last we have traversed the right subtree by passing root->right as a parameter.

Number of leaf nodes in first Tree are 4 Number of leaf nodes in second tree are 1 Number of leaf nodes in third tree are 1

**Sanfoundry Global Education & Learning Series – 1000 C++ Programs.**

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