Reversal order of linked list using recursion in swift

Swift program for Reversal order of linked list using recursion. Here problem description and other solutions.

import Foundation
// Swift 4 program for
// Print reverse of a linked list without actually reversing

// Linked list node
class LinkNode
{
	var data: Int;
	var next: LinkNode? ;
	init(_ data: Int)
	{
		self.data = data;
		self.next = nil;
	}
}
class SingleLL
{
	var head: LinkNode? ;
	var tail: LinkNode? ;
	init()
	{
		self.head = nil;
		self.tail = nil;
	}
	// Add new Node at end of linked list 
	func addNode(_ data: Int)
	{
		let node: LinkNode? = LinkNode(data);
		if (self.head == nil)
		{
			self.head = node;
		}
		else
		{
			// Append the node at last position
			self.tail!.next = node;
		}
		self.tail = node;
	}
	// Display reversal view of linked list using recursion
	func printReverse(_ node: LinkNode? )
	{
		if (node == nil)
		{
			return;
		}
		// Visit to next node
		self.printReverse(node!.next);
		// Display node value
		print("  "
			+ String(node!.data), terminator: "");
	}
	static func main()
	{
		let sll: SingleLL = SingleLL();
		//  1 → 2 → 8 → 4 → 9 → 6 → NULL
		sll.addNode(1);
		sll.addNode(2);
		sll.addNode(8);
		sll.addNode(4);
		sll.addNode(9);
		sll.addNode(6);
		// Reversal view
		// 6 9 4 8 2 1
		sll.printReverse(sll.head);
	}
}
SingleLL.main();

Output

  6  9  4  8  2  1