【LeetCode 230】Kth Smallest Element in a BST

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.

 

题意：

　　给定一个二分搜索树，返回第K小的结点

思路：

　　只要明白BST树的原理，只要中序遍历一遍BST树即可。求第K小的，只需遍历前K个结点就OK。

C++：

 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 

13     int ret, cnt, _k;

14 

15     void rec(TreeNode *root)

16     {

17         if(root->left == 0 && root->right == 0)

18         {

19             cnt++;

20             if(cnt == _k)

21                 ret = root->val;

22             return ;

23         }

24         

25         if(root->left != 0)

26             rec(root->left);

27         

28         cnt++;

29         if(cnt == _k){

30             ret = root->val;

31             return ;

32         }

33         

34         if(root->right != 0)

35             rec(root->right);

36     }

37     

38     int kthSmallest(TreeNode* root, int k) {

39         if(root == 0)

40             return 0;

41             

42         _k = k; cnt = ret = 0;

43         

44         rec(root);

45         

46         return ret;

47     }

48 };

 

Python：

 1 # Definition for a binary tree node.

 2 # class TreeNode:

 3 #     def __init__(self, x):

 4 #         self.val = x

 5 #         self.left = None

 6 #         self.right = None

 7 

 8 class Solution:

 9     # @param {TreeNode} root

10     # @param {integer} k

11     # @return {integer}

12     

13     def __init__(self):

14         self.cnt = 0

15         self.ret = 0

16         

17     def rec(self, root, k):

18         if root.left is None and root.right is None:

19             self.cnt = self.cnt + 1

20             if self.cnt == k:

21                 self.ret = root.val

22             return 

23         

24         if root.left is not None:

25             self.rec(root.left, k)

26             

27         self.cnt = self.cnt + 1

28         if self.cnt == k:

29             self.ret = root.val

30             return 

31         

32         if root.right is not None:

33             self.rec(root.right, k)

34         

35     def kthSmallest(self, root, k):

36         if root is None:

37             return 0

38         

39         self.rec(root, k)

40 

41         return self.ret