【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++：

 /**
  * 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 ret, cnt, _k;

     void rec(TreeNode *root)
     {
         if(root->left ==  && root->right == )
         {
             cnt++;
             if(cnt == _k)
                 ret = root->val;
             return ;
         }

         if(root->left != )
             rec(root->left);

         cnt++;
         if(cnt == _k){
             ret = root->val;
             return ;
         }

         if(root->right != )
             rec(root->right);
     }

     int kthSmallest(TreeNode* root, int k) {
         if(root == )
             return ;

         _k = k; cnt = ret = ;

         rec(root);

         return ret;
     }
 };

Python：

 # Definition for a binary tree node.
 # class TreeNode:
 #     def __init__(self, x):
 #         self.val = x
 #         self.left = None
 #         self.right = None

 class Solution:
     # @param {TreeNode} root
     # @param {integer} k
     # @return {integer}

     def __init__(self):
         self.cnt = 0
         self.ret = 0

     def rec(self, root, k):
         if root.left is None and root.right is None:
             self.cnt = self.cnt + 1
             if self.cnt == k:
                 self.ret = root.val
             return 

         if root.left is not None:
             self.rec(root.left, k)

         self.cnt = self.cnt + 1
         if self.cnt == k:
             self.ret = root.val
             return 

         if root.right is not None:
             self.rec(root.right, k)

     def kthSmallest(self, root, k):
         if root is None:
             return 0

         self.rec(root, k)

         return self.ret