X = 0;
	for i = 1 to N do
	-     for j = i to N do
	-     for k = j to N do
	-     X++;
	
 
 
H --> (a) --> (b) --> (c) --> (d) --> (e) --> null.Write a Java method called LastButOne to return a reference to the last but one node (this is node (d) in the figure above). In case the list has only one node, then return H itself. Here is the header for your method:
	public Node LastButOne(Node H);
	pre: (e, c, a, b, d)Draw the binary tree. HINT: looking at the pre-order list, you know where to place node e and node c. But node a has two possible places.
post: (a, b, c, d, e)
	class Hello {
	  public static void main(String[] args){
		System.out.println("Hello World!");
	  }
	}
	
	public Node LastButOne(Node H){
	   Node Prev = H;
	   H = H.getNext();
	   if (H == null) {
		return Prev;
	   };
	   while (H.getNext() != null){
		Prev = H;
		H = H.getNext();
	   }
	}