WAP to find Kth Maximum or Largest in Binar Search Tree (BST)

class BST  

{
 private static Node root;
  
 public static void main(String [] args) {
  int [] A = {8, 3, 10, 1, 6, 14, 4, 7, 13};
  root = buildBST(root, A);
  System.out.println("inorder: ");
  printBSTInorder(root);
  System.out.println();
  int sizeBST = sizeOfBST(root);
  System.out.println("Size of BST: " + sizeBST);
   
  for (int i=1; i<=sizeBST; i++) {
   Node node = nthLargestNode(root, i);
   System.out.println(i + " largest node: " + node.data);
  }
 }
  
 public static Node nthLargestNode(Node node, int N) {
   if (null == node) 
  {
   return null;
  }
    
  int R = sizeOfBST(node.right) + 1;
  if (N == R) 
  {
   return node;
  }
  if(N < R) 
  {
   return nthLargestNode(node.right, N);
  } 
  else 
  {
   return nthLargestNode(node.right, N-R);
  }
  }
 
  
 public static int sizeOfBST(Node node) {
  if (null == node) { return 0; }
  return (sizeOfBST(node.left) + 1 + sizeOfBST(node.right));
 }
  
 public static Node buildBST(Node node, int [] A) {
  if (A == null) { return null; }
  int len = A.length;
  for (int i=0; i<len; i++) {
   node = insertIntoBST(node, A[i]);
  }
  return node;
 }
  
 private static Node insertIntoBST(Node node, int value) {
  if (null == node) {
   return new Node(value);
  }
  if (value <= node.data) {
   node.left = insertIntoBST(node.left, value);
  } else {
   node.right = insertIntoBST(node.right, value);
  }
  return node;
 }
  
 public static void printBSTInorder(Node node) {
  if (null == node) {  return ; }
  if (null != node.left) {
   printBSTInorder(node.left);
  }
  System.out.print(node.data + " ");
  if (null != node.right) {
   printBSTInorder(node.right);
  }
 }
}
 
class Node {
 Integer data;
 Node left;
 Node right;
  
 public Node(Integer data) {
  this.data = data;
  this.left = null;
  this.right = null;
 }
}

 Run Here http://ideone.com/rUQbn

 Time Complexity O(N^2) N is number of nodes in BST

 Space Complexity O(1)   Ping me if anything wrong here