230. Kth Smallest Element in a BST

Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.

Note:
You may assume k is always valid, 1 ≤ k ≤ BST's total elements.

Follow up:
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?

题目含义：给出二叉搜索树中第k小的数字

 1     private int leftTreeNodeCount(TreeNode root) {
 2         if (root == null) return 0;
 3         return 1 + leftTreeNodeCount(root.left) + leftTreeNodeCount(root.right);
 4     }
 5     
 6     public int kthSmallest(TreeNode root, int k) {
 7 //        在二叉搜索树种，找到第K个小的元素。
 8 //        算法如下：
 9 //        1、计算左子树元素个数left。
10 //        2、 left+1 = K，则根节点即为第K个元素
11 //        3、left >=k, 则第K个元素在左子树中，
12 //        4、left +1 <k, 则转换为在右子树中，寻找第K-left-1元素
13         int leftCount = leftTreeNodeCount(root.left);
14         if (leftCount >= k) {
15             return kthSmallest(root.left, k);
16         } 
17         if (leftCount + 1 < k) return kthSmallest(root.right, k - leftCount - 1);
18         return root.val;        
19     }