題目：
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?
解答：
《編程之美》中有講過類似的題目：尋找數組（未排序）中的第K大數。這裡是BST，已經有順序，更方便一些。
如果左子樹數目大於等於K，那麼直接找左子樹中第K個；如果左子樹數目等於 K-1，那麼根節點就是第K個；如果左子樹數目小於K，那麼直接找右子樹中第（K - leftSum - 1）個（勿忘根節點） 關於左子樹有多少結點？需要另一個遞歸程序計算。
/**
* 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 findNodeSum(TreeNode* root) {
if (root == NULL) return 0;
return (findNodeSum(root->left) + findNodeSum(root->right) + 1);
}
int kthSmallest(TreeNode* root, int k) {
int leftSum = findNodeSum(root->left);
if (leftSum >= k) return kthSmallest(root->left, k);
else if (leftSum == k - 1) return root->val;
else
{
return kthSmallest(root->right,k - leftSum - 1);
}
}
};
題目到這兒還沒完，如果BST總是在修改呢？
最好應該修改樹結點的結構，增加一項屬性：左子樹的節點總數。這樣搜索的復雜度就是 O (height)
