From 0b3945cfd1a9b972839678b717f0f08461daae34 Mon Sep 17 00:00:00 2001
From: jojoo15 <75017412+jojoo15@users.noreply.github.com>
Date: Tue, 18 May 2021 12:18:00 +0200
Subject: [PATCH] =?UTF-8?q?Update=200700.=E4=BA=8C=E5=8F=89=E6=90=9C?=
 =?UTF-8?q?=E7=B4=A2=E6=A0=91=E4=B8=AD=E7=9A=84=E6=90=9C=E7=B4=A2.md?=
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recursive python version of code modified
---
 problems/0700.二叉搜索树中的搜索.md | 17 +++++++++++------
 1 file changed, 11 insertions(+), 6 deletions(-)

diff --git a/problems/0700.二叉搜索树中的搜索.md b/problems/0700.二叉搜索树中的搜索.md
index 277ef681..3e35fbfb 100644
--- a/problems/0700.二叉搜索树中的搜索.md
+++ b/problems/0700.二叉搜索树中的搜索.md
@@ -212,13 +212,18 @@ Python：
 递归法：
 
 ```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
 class Solution:
     def searchBST(self, root: TreeNode, val: int) -> TreeNode:
-        if root is None:
-            return None
-        if val < root.val: return self.searchBST(root.left, val)
-        elif val > root.val: return self.searchBST(root.right, val)
-        else: return root
+        if not root or root.val == val: return root  //为空或者已经找到都是直接返回root，所以合并了
+        if root.val > val: return self.searchBST(root.left,val)  //注意一定要加return
+        else: return self.searchBST(root.right,val)  
+
 ```
 
 迭代法：
@@ -243,4 +248,4 @@ Go：
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
 * B站视频：[代码随想录](https://space.bilibili.com/525438321)
 * 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-<div align="center"><img src=../pics/公众号.png width=450 alt=> </img></div>
\ No newline at end of file
+<div align="center"><img src=../pics/公众号.png width=450 alt=> </img></div>