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?

Analyse:

Trial 1: We can first sort the sequence by inorder traversal and find the k-th element.

　　　　Runtime: 24ms

/**
 * 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) {
        if(!root) return 0;
        vector<int> result;
        inorder(root, result);
        return result[k - 1];
    }
    void inorder(TreeNode* root, vector<int>& result){
        if(!root) return;
        inorder(root->left, result);
        result.push_back(root->val);
        inorder(root->right, result);
    }
};

Trial 2: Do the inorder process iterately, set a count number, when it equals to k, return corresponding value.

        　Runtime: 24ms.

/**
 * 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) {
        if(!root) return 0;
        stack<TreeNode* > stk;
        int order = 0;
        while(!stk.empty() || root){
            if(root){
                stk.push(root);
                root = root->left;
            }
            else{
                root = stk.top();
                stk.pop();
                order++;
                if(order == k) return root->val;
                root = root->right;
            }
        }
    }
};