[leedcode 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?

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
  /*  两种思路： 
        1.空间换时间 
        BST的特性是，如果按照中序排列，得到的递增序；所以可以使用一个stack进行中序遍历，直到找到第K个元素； 
        2. 树树的结点数 
        对于每个节点root，计算以它为根节点的左子树的节点数，计作S。 
        如果S+1==K，返回root->val; 
        如果S+1 > K，在root的左子树里面查找第K小元素； 
        如果S+1 > k ,在root的右子树里面查找第k-s-1小元素*/
    public int kthSmallest(TreeNode root, int k) {
        /*Stack<TreeNode> stack=new Stack<TreeNode>();
        while(!stack.isEmpty()||root!=null){
            while(root!=null){
                stack.push(root);
                root=root.left;
            }
            TreeNode temp=stack.pop();
            k--;
            if(k==0){
                return temp.val;
            }
            root=temp.right;
        }
        return 0;*/
        if(root==null) return 0;
        int left=find(root.left);
        if(left+1==k) return root.val;
        if(left+1<k) return kthSmallest(root.right,k-left-1);
        else return kthSmallest(root.left,k);

    }
    public int find(TreeNode p){
        if(p==null) return 0;
        return find(p.left)+find(p.right)+1;
    }
}