Count complete odd nodes paths of binary tree in java

Java program for Count complete odd nodes paths of binary tree. Here mentioned other language solution.

/* 
  Java program for
  Count odd paths in Binary Tree
*/
// Binary Tree node
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;
	}
}
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 countOddNodePath(node.left) + 
              countOddNodePath(node.right);
		}
	}
	public static void main(String[] args)
	{
		// New of binary tree
		BinaryTree 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.
		System.out.println(tree.countOddNodePath(tree.root));
	}
}

Output

3