【LeetCode从零单刷】Kth Smallest Element in a BST

题目：

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：

《编程之美》中有讲过类似的题目：寻找数组（未排序）中的第K大数。这里是BST，已经有顺序，更方便一些。

1. 如果左子树数目大于等于K，那么直接找左子树中第K个；
2. 如果左子树数目等于 K-1，那么根节点就是第K个；
3. 如果左子树数目小于K，那么直接找右子树中第（K - leftSum - 1）个（勿忘根节点）

/**
 * 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);
        }
    }
};

最好应该修改树结点的结构，增加一项属性：左子树的节点总数。这样搜索的复杂度就是 O (height)

【LeetCode从零单刷】Kth Smallest Element in a BST

原文：http://blog.csdn.net/ironyoung/article/details/49507419