From 970a441f1034e7c65881c4867d03de300776589f Mon Sep 17 00:00:00 2001
From: Asterisk <44215173+casnz1601@users.noreply.github.com>
Date: Thu, 7 Oct 2021 19:29:26 +0800
Subject: [PATCH] =?UTF-8?q?Update=200501.=E4=BA=8C=E5=8F=89=E6=90=9C?=
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规范Python3代码
---
 problems/0501.二叉搜索树中的众数.md | 68 ++++++++++++--------
 1 file changed, 42 insertions(+), 26 deletions(-)

diff --git a/problems/0501.二叉搜索树中的众数.md b/problems/0501.二叉搜索树中的众数.md
index 29e5e139..18d9b290 100644
--- a/problems/0501.二叉搜索树中的众数.md
+++ b/problems/0501.二叉搜索树中的众数.md
@@ -470,38 +470,54 @@ class Solution {
 
 ## Python 
 
-递归法
+> 递归法
 
-```python
+```python3
+# 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 __init__(self):
+        self.pre = TreeNode()
+        self.count = 0
+        self.max_count = 0
+        self.result = []
+
     def findMode(self, root: TreeNode) -> List[int]:
-        if not root: return
-        self.pre = root
-        self.count = 0   //统计频率
-        self.countMax = 0  //最大频率
-        self.res = []
-        def findNumber(root):
-            if not root: return None  // 第一个节点
-            findNumber(root.left)  //左
-            if self.pre.val == root.val:  //中: 与前一个节点数值相同
-                self.count += 1
-            else:  // 与前一个节点数值不同
-                self.pre = root
-                self.count = 1
-            if self.count > self.countMax:  // 如果计数大于最大值频率
-                self.countMax = self.count  // 更新最大频率
-                self.res = [root.val]  //更新res
-            elif self.count == self.countMax:  // 如果和最大值相同，放进res中
-                self.res.append(root.val)
-            findNumber(root.right)  //右
-            return
-        findNumber(root)
-        return self.res
+        if not root: return None
+        self.search_BST(root)
+        return self.result
+        
+    def search_BST(self, cur: TreeNode) -> None:
+        if not cur: return None
+        self.search_BST(cur.left)
+        # 第一个节点
+        if not self.pre:
+            self.count = 1
+        # 与前一个节点数值相同
+        elif self.pre.val == cur.val:
+            self.count += 1 
+        # 与前一个节点数值不相同
+        else:
+            self.count = 1
+        self.pre = cur
+
+        if self.count == self.max_count:
+            self.result.append(cur.val)
+        
+        if self.count > self.max_count:
+            self.max_count = self.count
+            self.result = [cur.val]	# 清空self.result，确保result之前的的元素都失效
+        
+        self.search_BST(cur.right)
 ```
 	
 
-迭代法-中序遍历-不使用额外空间，利用二叉搜索树特性 
-```python
+> 迭代法-中序遍历-不使用额外空间，利用二叉搜索树特性 
+```python3
 class Solution:
     def findMode(self, root: TreeNode) -> List[int]:
         stack = []