[LeetCode-230]Kth Smallest Element in a BST

本文介绍了一种方法来找到给定二叉搜索树中的第k小元素,包括基本实现和针对频繁修改的优化策略。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

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?

Solution:

/**
 * 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 tree_size(TreeNode* root){
        if (root){
            return tree_size(root->left) + tree_size(root->right) + 1;
        } else {
            return 0;
        }
    }
    int kthSmallest(TreeNode* root, int k) {
        if (tree_size(root->left) + 1 == k) {
            return root->val;
        } else if (tree_size(root->left) + 1 > k) {
            return kthSmallest(root->left, k);
        } else {
            return kthSmallest(root->right, k-(tree_size(root->left)+1));
        }
    }
};
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值