Triple order traversal of a binary tree in kotlin

Kotlin program for Triple order traversal of a binary tree. Here mentioned other language solution.

/*
    Kotlin program for
    Triple order traversal of a binary tree
*/
// 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 tripleOrder(node: TreeNode ? ): Unit
	{
		if (node != null)
		{
			// Print preorder node
			print(" " + node.data);
			// Visit left subtree
			this.tripleOrder(node.left);
			// Print inorder node
			print(" " + node.data);
			// Visit right subtree
			this.tripleOrder(node.right);
			// Print postorder node
			print(" " + node.data);
		}
	}
}
fun main(args: Array < String > ): Unit
{
	// Create new binary trees
	val tree: BinaryTree = BinaryTree();
	/*
	         4    
	        / \                         
	       /   \    
	     -4     7    
	     / \     \               
	    2   3     1
	       / \   / 
	      6   8 5
	     /       
	    9          
	  ----------------------
	  Constructing binary tree
	        
	*/
	tree.root = TreeNode(4);
	tree.root?.left = TreeNode(-4);
	tree.root?.left?.right = TreeNode(3);
	tree.root?.left?.right?.left = TreeNode(6);
	tree.root?.left?.right?.left?.left = TreeNode(9);
	tree.root?.left?.right?.right = TreeNode(8);
	tree.root?.left?.left = TreeNode(2);
	tree.root?.right = TreeNode(7);
	tree.root?.right?.right = TreeNode(1);
	tree.root?.right?.right?.left = TreeNode(5);
	// Print triple order
	// ----------------------------------
	// 4 -4 2 2 2 -4 3 6 9 9 9 6 6 3 8 8 ..
	// 8 3 -4 4 7 7 1 5 5 5 1 1 7 4
	tree.tripleOrder(tree.root);
}

Output

 4 -4 2 2 2 -4 3 6 9 9 9 6 6 3 8 8 8 3 -4 4 7 7 1 5 5 5 1 1 7 4