Update 230._kth_smallest_element_in_a_bst.md

0xkookoo · web-flow · commit adcd18bb6b8d · 2018-09-24T08:50:50.000-05:00
diff --git a/docs/Leetcode_Solutions/Python/230._kth_smallest_element_in_a_bst.md b/docs/Leetcode_Solutions/Python/230._kth_smallest_element_in_a_bst.md
@@ -1,11 +1,49 @@
-### 230. Kth Smallest Element in a BST
+# 230. Kth Smallest Element in a BST
 
-题目:
-<https://leetcode.com/problems/kth-smallest-element-in-a-bst/>
+**<font color=red>难度: Medium</font>**
 
+## 刷题内容
 
-难度:
-Medium
+> 原题连接
+
+* https://leetcode.com/problems/kth-smallest-element-in-a-bst/description/
+
+> 内容描述
+
+```
+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.
+
+Example 1:
+
+Input: root = [3,1,4,null,2], k = 1
+   3
+  / \
+ 1   4
+  \
+   2
+Output: 1
+Example 2:
+
+Input: root = [5,3,6,2,4,null,null,1], k = 3
+       5
+      / \
+     3   6
+    / \
+   2   4
+  /
+ 1
+Output: 3
+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?
+```
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(N)******- 空间复杂度: O(N)******
 
 
 跟昨天做的一道题类似，一上来就走取巧之路。
@@ -35,7 +73,7 @@ class Solution(object):
 ```
 
 
-现在看到kth 就条件反射的想用divide & conquer, 扫root的左子树看nodes量，如果nodes数量是k-1，那么node就刚好是第k个，如果大于k > 左子树数量，扫右子树，同时更新root为root.right。
+现在看到kth 就条件反射的想用divide & conquer, 扫root的左子树看nodes量，如果nodes数量是k-1，那么root就刚好是第k个，如果大于k > 左子树数量，扫右子树，同时更新root为root.right。
 
 看到的言论：
 
