LeetCode Kth Smallest Element in a BST

最新推荐文章于 2025-08-18 21:25:39 发布

转载最新推荐文章于 2025-08-18 21:25:39 发布 · 44 阅读

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原文链接：http://www.cnblogs.com/lailailai/p/4619101.html

文章标签：

#数据结构与算法 #runtime

本文介绍了一种高效的方法来找到二叉搜索树中的第K小元素，利用了二叉搜索树的性质进行中序遍历，并通过在树节点中添加一个count字段来动态计算排名，实现O(height of BST)的复杂度。

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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?

Hint:

Try to utilize the property of a BST.
What if you could modify the BST node's structure?
The optimal runtime complexity is O(height of BST).
Credits:
Special thanks to @ts for adding this problem and creating all test cases.

如果树节点结构是这样的话，直接进行中序遍历得出，如果在树中的节点添加一个count字段用于表示该子树含有的数值数量就可以在logn的时间里得出。就是一个动态计算rank的功能。

下面是inorder的代码：

/**
 * 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 {
private:
    int idx;
    int value;
    int K;
public:
    int kthSmallest(TreeNode* root, int k) {
        K = k;
        idx = 0;
        dfs(root);
        return value;
    }
    
    void dfs(TreeNode* root) {
        if (root == NULL) {
            return;
        }
        dfs(root->left);
        idx++;
        if (idx == K) {
            value = root->val;
            return;
        }
        dfs(root->right);
    }
};

转载于:https://www.cnblogs.com/lailailai/p/4619101.html