Check for foldable binary tree in kotlin

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

/* 
  Kotlin 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 ? ;
	constructor(data: Int)
	{
		// Set node value
		this.data = data;
		this.left = null;
		this.right = null;
	}
}
class BinaryTree
{
	var root: TreeNode ? ;
	constructor()
	{
		this.root = null;
	}
	fun isFoldable(leftRoot: TreeNode ? , 
                   rightRoot : TreeNode ? ): Boolean
	{
		if (leftRoot == null && rightRoot == null)
		{
			// When both node empty
			return true;
		}
		else if (leftRoot != null && rightRoot != null &&
                 this.isFoldable(leftRoot.left, rightRoot.right) && 
                 this.isFoldable(leftRoot.right, rightRoot.left))
		{
			// When
			// Valid the properties of foldable tree
			return true;
		}
		// Fail case
		return false;
	}
	// Display tree element inorder form
	fun inorder(node: TreeNode ? ): Unit
	{
		if (node != null)
		{
			// Visit to left subtree
			this.inorder(node.left);
			// Print node value
			print("  " + node.data);
			// Visit to right subtree
			this.inorder(node.right);
		}
	}
}
fun main(args: Array < String > ): Unit
{
	// Create new tree
	val 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: Boolean = tree.isFoldable(tree.root?.left, 
                                          tree.root?.right);
	if (result == true)
	{
		println("\n Foldable Tree");
	}
	else
	{
		println("\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)
	{
		println("\n Foldable Tree");
	}
	else
	{
		println("\n Not Foldable Tree");
	}
}

Output

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