Count complete odd nodes path of binary tree in typescript

Ts program for Count complete odd nodes path of binary tree. Here mentioned other language solution.

/* 
  TypeScript program for
  Count odd paths in Binary Tree
*/
// Binary Tree node
class TreeNode
{
	public data: number;
	public left: TreeNode;
	public right: TreeNode;
	constructor(data: number)
	{
		// Set node value
		this.data = data;
		this.left = null;
		this.right = null;
	}
}
class BinaryTree
{
	public root: TreeNode;
	constructor()
	{
		// Set initial tree root
		this.root = null;
	}
	// Count all paths from root to leaf
	// which containing all Odd nodes
	public number countOddNodePath(node: TreeNode)
	{
		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 main()
	{
		// 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.log(tree.countOddNodePath(tree.root));
	}
}
BinaryTree.main();
/*
 file : code.ts
 tsc --target es6 code.ts
 node code.js
 */

Output

3