Program to find count of nodes having 1 child

// C++ implementation to print
// the nodes having a single child
#include <bits/stdc++.h>
using namespace std;

// Class of the Binary Tree node
struct Node
{
int data;
Node *left, *right;

Node(int x)
{
	data = x;
	left = right = NULL;
}
};

vector<int> lst;

// Function to find the nodes
// having single child
void printNodesOneChild(Node* root)
{
// Base case
if (root == NULL)
	return;

// Condition to check if the
// node is having only one child
if (root->left != NULL &&
	root->right == NULL ||
	root->left == NULL &&
	root->right != NULL)
{
	lst.push_back(root->data);
}

// Traversing the left child
printNodesOneChild(root -> left);

// Traversing the right child
printNodesOneChild(root -> right);
}

//Driver code
int main()
{
// Constructing the binary tree
Node *root = new Node(2);
root -> left = new Node(3);
root -> right = new Node(5);
root -> left -> left = new Node(7);
root -> right -> left = new Node(8);
root -> right -> right = new Node(6);

// Function calling
printNodesOneChild(root);

// Condition to check if there is
// no such node having single child
if (lst.size() == 0)
	printf("-1");
else
{
	for(int value : lst)
	{
	cout << (value) << endl;
	}
}
}