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.

兩種思路：
1.空間換時間
BST的特性是，如果按照中序排列，得到的遞增序；所以可以使用一個stack進行中序遍歷，直到找到第K個元素；
2. 樹樹的結點數
對於每個節點root，計算以它為根節點的樹的節點數，計作S。
如果S==K，返回root->val;
如果S > K，在root的左子樹裡面查詢第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<TreeNode> 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程式碼：

/**
 * 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<k){
            return kthSmallest(root->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;
    }
};