Count complete odd nodes path of binary tree in swift

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

import Foundation
/* 
  Swift 4 program for
  Count odd paths in Binary Tree
*/
// Binary Tree node
class TreeNode
{
	var data: Int;
	var left: TreeNode? ;
	var right: TreeNode? ;
	init(_ data: Int)
	{
		// Set node value
		self.data = data;
		self.left = nil;
		self.right = nil;
	}
}
class BinaryTree
{
	var root: TreeNode? ;
	init()
	{
		self.root = nil;
	}
	// Count all paths from root to leaf
	// which containing all Odd nodes
	func countOddNodePath(_ node: TreeNode? ) -> Int
	{
		if (node == nil || node!.data % 2 == 0)
		{
			// When tree node is null or
			// its contain even value
			return 0;
		}
		else
		{
			if (node!.left == nil && node!.right == nil)
			{
				// When get leaf node And
				// path contain all Odd nodes
				return 1;
			}
			return self.countOddNodePath(node!.left) + 
              self.countOddNodePath(node!.right);
		}
	}
	static func main()
	{
		// New of binary tree
		let tree: BinaryTree = BinaryTree();
		/*
		  Construct Binary Tree
		  -----------------------
		         5
		        /  \ 
		       /    \
		      3      9
		     / \    / \
		    7   6  1   3
		   / \    / \   
		  10  3  7   4
		*/
		// Add tree node
		tree.root = TreeNode(5);
		tree.root!.left = TreeNode(3);
		tree.root!.right = TreeNode(9);
		tree.root!.right!.right = TreeNode(3);
		tree.root!.left!.right = TreeNode(6);
		tree.root!.right!.left = TreeNode(1);
		tree.root!.left!.left = TreeNode(7);
		tree.root!.left!.left!.left = TreeNode(10);
		tree.root!.left!.left!.right = TreeNode(3);
		tree.root!.right!.left!.right = TreeNode(4);
		tree.root!.right!.left!.left = 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.
		print(tree.countOddNodePath(tree.root));
	}
}
BinaryTree.main();

Output

3