LeetCode #230 - Kth Smallest Element in a BST

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本文介绍了一种在二叉搜索树(BST)中寻找第K小元素的有效算法。通过计算左子树节点数判断目标节点位置，实现快速查找。

题目描述：

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?

求BST的第k小的值，先计算左子树的节点数，如果数量大于等于k说明目标在左子树，如果等于k-1，说明根节点即为所求，否则说明目标值在右子树。

class Solution {
public:
    int kthSmallest(TreeNode* root, int k) {
        int left_num=countNode(root->left);
        if(left_num==k-1) return root->val;
        else if(left_num>=k) return kthSmallest(root->left,k);
        else if(left_num<k-1) return kthSmallest(root->right,k-left_num-1);
    }
    
    int countNode(TreeNode* root)
    {
        if(root==NULL) return 0;
        return 1+countNode(root->left)+countNode(root->right);
    }
};