Check for foldable binary tree in swift

Swift program for Check for foldable binary tree. Here problem description and other solutions.

import Foundation
/* 
  Swift 4 program for
  Check if binary tree is foldable binary tree
  Using recursion
*/
// 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;
	}
	func isFoldable(_ leftRoot: TreeNode? , 
                    _ rightRoot : TreeNode? ) -> Bool
	{
		if (leftRoot == nil && rightRoot == nil)
		{
			// When both node empty
			return true;
		}
		else if (leftRoot  != nil && rightRoot  != nil &&
                 self.isFoldable(leftRoot!.left, rightRoot!.right) &&
                 self.isFoldable(leftRoot!.right, rightRoot!.left))
		{
			// When
			// Valid the properties of foldable tree
			return true;
		}
		// Fail case
		return false;
	}
	// Display tree element inorder form
	func inorder(_ node: TreeNode? )
	{
		if (node  != nil)
		{
			// Visit to left subtree
			self.inorder(node!.left);
			// Print node value
			print("",node!.data, terminator: " ");
			// Visit to right subtree
			self.inorder(node!.right);
		}
	}
	static func main()
	{
		// Create new tree
		let tree: BinaryTree = BinaryTree();
		/*
		   Binary Tree
		  -----------------------
		       5
		     /   \
		    7     4
		   /       \
		  1         2
		   \       /
		    2     5
		*/
		// Binary tree nodes
		tree.root = TreeNode(5);
		tree.root!.left = TreeNode(7);
		tree.root!.right = TreeNode(4);
		tree.root!.left!.left = TreeNode(1);
		tree.root!.right!.right = TreeNode(2);
		tree.root!.right!.right!.left = TreeNode(5);
		tree.root!.left!.left!.right = TreeNode(2);
		// Display Tree elements
		tree.inorder(tree.root);
		var result: Bool = tree.isFoldable(
          tree.root!.left, tree.root!.right);
		if (result == true)
		{
			print("\n Foldable Tree");
		}
		else
		{
			print("\n Not Foldable Tree");
		}
		/*
		         5
		       /   \
		      7     4
		     /       \
		    1         2
		     \       /
		      2     5
		       \
		        3  
		*/
		// Case 2
		tree.root!.left!.left!.right!.right = TreeNode(3);
		tree.inorder(tree.root);
		result = tree.isFoldable(
          tree.root!.left, tree.root!.right);
		if (result == true)
		{
			print("\n Foldable Tree");
		}
		else
		{
			print("\n Not Foldable Tree");
		}
	}
}
BinaryTree.main();

Output

 1  2  7  5  4  5  2
 Foldable Tree
 1  2  3  7  5  4  5  2
 Not Foldable Tree