Give an algorithm to find the number of leaves in a binary tree.

Question: Given the root pointer to a binary tree, find the number of leaves.

Input: Sample Tree (Pointer to node 1 is given). Find the number of leaves.
Output: Number of leaves = 4

According to the definition, the set of all nodes whose both left and right child are null, are known as leaf nodes.

In the above example, we have the nodes – 7, 9, 5, 4 as the leaf nodes.

We can perform a level order traversal and count the number of nodes, who have their right child and left child as null. Return the count in the end.

#include<stdio.h>
#include<malloc.h>

struct binaryTreeNode{
	int data;
	struct binaryTreeNode * left;
	struct binaryTreeNode * right;
};

int numberOfLeaves(struct binaryTreeNode * root)
{
	// Level order traversal
	struct binaryTreeNode * temp = NULL;
	struct queue * Q = NULL;
	
	// Maintain a count
	int count = 0;

	if(root == NULL)
		return 0;

	Q = enQueue(Q, root);
	while(!isQueueEmpty(Q))
	{
		temp = Q -> front -> data;
		
		// Now check if the node is a leaf node
		if(temp -> left == NULL && temp -> right == NULL)
		{
			// This means a leaf node
			count++;
		}
		
		Q = deQueue(Q);
		if(temp -> left)
			Q = enQueue(Q, temp -> left);
		if(temp -> right)
			Q = enQueue(Q, temp -> right);
	}
	// Delete the queue
	free(Q);
	
	// Now return the count
	return count;
}

Time Complexity:- O(n)
Space Complexity:- O(n)

Ideone link for the sample program:- http://ideone.com/umTait