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?

Hint:

Try to utilize the property of a BST.Show More Hint
Credits:
Special thanks to @ts for adding this problem and creating all test cases.

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树的节点数，计作S。
如果S==K，返回root->val;
如果S > K，在root的左子树里面查找第K小元素；
如果S

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int kthSmallest(TreeNode root, int k) {
        Stack
   
     s = new Stack<>(); TreeNode p=root; s.push(p); int cnt=0; do{ while(p!=null){ s.push(p); p=p.left; } TreeNode top=s.pop(); cnt++; if(cnt==k){ return top.val; } p=top.right; }while(!s.empty()); return 0; } }
   

思路2代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int kthSmallest(TreeNode* root, int k) {
        int left=findNodesSum(root->left);
        if(left+1==k){
            return root->val;
        }else if(left+1
   
    right,k-left-1); }else{ return kthSmallest(root->left,k); } } int findNodesSum(TreeNode* root){ if(!root){ return 0; } int sum = findNodesSum(root->left)+findNodesSum(root->right)+1; return sum; } };
   

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