Count complete odd nodes path of binary tree in c#

Csharp program for Count complete odd nodes path of binary tree. Here more solutions.

// Include namespace system
using System;
/* 
  Csharp program for
  Count odd paths in Binary Tree
*/
// Binary Tree node
public class TreeNode
{
	public int data;
	public TreeNode left;
	public TreeNode right;
	public TreeNode(int data)
	{
		// Set node value
		this.data = data;
		this.left = null;
		this.right = null;
	}
}
public class BinaryTree
{
	public TreeNode root;
	public BinaryTree()
	{
		//Set initial tree root
		this.root = null;
	}
	// Count all paths from root to leaf
	// which containing all Odd nodes
	public int countOddNodePath(TreeNode node)
	{
		if (node == null || node.data % 2 == 0)
		{
			// When tree node is null or
			// its contain even value
			return 0;
		}
		else
		{
			if (node.left == null && node.right == null)
			{
				// When get leaf node And
				// path contain all Odd nodes
				return 1;
			}
			return this.countOddNodePath(node.left) +
              this.countOddNodePath(node.right);
		}
	}
	public static void Main(String[] args)
	{
		// New of binary tree
		var tree = new BinaryTree();
		/*
		  Construct Binary Tree
		  -----------------------
		         5
		        /  \ 
		       /    \
		      3      9
		     / \    / \
		    7   6  1   3
		   / \    / \   
		  10  3  7   4
		*/
		// Add tree node
		tree.root = new TreeNode(5);
		tree.root.left = new TreeNode(3);
		tree.root.right = new TreeNode(9);
		tree.root.right.right = new TreeNode(3);
		tree.root.left.right = new TreeNode(6);
		tree.root.right.left = new TreeNode(1);
		tree.root.left.left = new TreeNode(7);
		tree.root.left.left.left = new TreeNode(10);
		tree.root.left.left.right = new TreeNode(3);
		tree.root.right.left.right = new TreeNode(4);
		tree.root.right.left.left = new TreeNode(7);
		/*
		  Given Binary Tree
		  -----------------------
		         5
		        /  \ 
		       /    \
		      3      9
		     / \    / \
		    7   6  1   3
		   / \    / \   
		  10  3  7   4
		  Here
		  Odd path from root to leaf
		  -----------------------
		         5
		        /  \ 
		       /    \
		      3      9
		     /      / \
		    7      1   3
		     \    /     
		      3  7    
		  ---------------------
			5->3->7->3    
			5->9->1->7  
		    5->9->3  
		  ----------------------
		  Result : 3 [number of path]
		*/
		// Count odd nodes paths from root
		// to leaf in a binary tree.
		Console.WriteLine(tree.countOddNodePath(tree.root));
	}
}

Output

3