Kth Smallest Element in a BST

最新推荐文章于 2025-09-10 21:59:39 发布

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原文链接：http://www.cnblogs.com/amazingzoe/p/4698615.html

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#runtime

本文介绍了一种高效的方法来找到二叉搜索树中第K小的元素，包括两种实现方式：排序遍历和迭代遍历。重点讨论了如何优化此操作以应对频繁修改的树结构。

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?

Analyse:

Trial 1: We can first sort the sequence by inorder traversal and find the k-th element.

　　　　Runtime: 24ms

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int kthSmallest(TreeNode* root, int k) {
13         if(!root) return 0;
14         vector<int> result;
15         inorder(root, result);
16         return result[k - 1];
17     }
18     void inorder(TreeNode* root, vector<int>& result){
19         if(!root) return;
20         inorder(root->left, result);
21         result.push_back(root->val);
22         inorder(root->right, result);
23     }
24 };

Trial 2: Do the inorder process iterately, set a count number, when it equals to k, return corresponding value.

　Runtime: 24ms.

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int kthSmallest(TreeNode* root, int k) {
13         if(!root) return 0;
14         stack<TreeNode* > stk;
15         int order = 0;
16         while(!stk.empty() || root){
17             if(root){
18                 stk.push(root);
19                 root = root->left;
20             }
21             else{
22                 root = stk.top();
23                 stk.pop();
24                 order++;
25                 if(order == k) return root->val;
26                 root = root->right;
27             }
28         }
29     }
30 };