LeetCode 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.

Example 1:

Input: root = [3,1,4,null,2], k = 1
   3
  / \
 1   4
  \
   2
Output: 1

Example 2:

Input: root = [5,3,6,2,4,null,null,1], k = 3
       5
      / \
     3   6
    / \
   2   4
  /
 1
Output: 3

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?

解析：从一颗二叉树中找到第K小的元素。考虑二叉树的性质，根节点的数值大于左子树，小于右子树。那么中序遍历获取到的数值是从小到大排序的，因此采用中序遍历，然后找到第K个元素即可，这里采用非递归的方式中序遍历，借助栈保存节点，当左子树和根节点遍历完了之后，出栈然后进入右子树。具体代码如下：

/**
 * 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) {
        TreeNode* p = root;
        stack<TreeNode*> s;
        while (!s.empty() || p)
        {
            while (p)
            {
                s.push(p);
                p = p->left;
            }
            if (!s.empty())
            {
                p = s.top();
                if(--k == 0)
                    return p->val;
                s.pop();
                p = p->right;
            }
        }
        return -1;
    }
};