From b87d619cdd1ab886055d94cb4b878f550fe51ea7 Mon Sep 17 00:00:00 2001
From: Kelvin <guanli611@outlook.com>
Date: Sun, 1 Aug 2021 13:54:51 -0400
Subject: [PATCH] =?UTF-8?q?Update=200098.=E9=AA=8C=E8=AF=81=E4=BA=8C?=
 =?UTF-8?q?=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91.md?=
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更改递归部分代码
1. 更新上一版代码格式 使其更符合PEP8.
2. 把代码细节都暴露出来, 逻辑更加清晰.
---
 problems/0098.验证二叉搜索树.md | 34 ++++++++++++++++++--------
 1 file changed, 24 insertions(+), 10 deletions(-)

diff --git a/problems/0098.验证二叉搜索树.md b/problems/0098.验证二叉搜索树.md
index 248d10f1..a271b977 100644
--- a/problems/0098.验证二叉搜索树.md
+++ b/problems/0098.验证二叉搜索树.md
@@ -337,6 +337,8 @@ class Solution {
 ```
 
 Python：
+
+**递归** - 利用BST中序遍历特性
 ```python
 # Definition for a binary tree node.
 # class TreeNode:
@@ -344,18 +346,30 @@ Python：
 #         self.val = val
 #         self.left = left
 #         self.right = right
-# 递归法
 class Solution:
     def isValidBST(self, root: TreeNode) -> bool:
-        res = []  //把二叉搜索树按中序遍历写成list
-        def buildalist(root):
-            if not root: return  
-            buildalist(root.left)  //左
-            res.append(root.val)  //中
-            buildalist(root.right)  //右
-            return res  
-        buildalist(root)
-        return res == sorted(res) and len(set(res)) == len(res) //检查list里的数有没有重复元素，以及是否按从小到大排列
+        # 思路: 利用BST中序遍历的特性.
+        # 中序遍历输出的二叉搜索树节点的数值是有序序列
+        candidate_list = []
+        
+        def __traverse(root: TreeNode) -> None: 
+            nonlocal candidate_list
+            if not root: 
+                return 
+            __traverse(root.left)
+            candidate_list.append(root.val)
+            __traverse(root.right)
+            
+        def __is_sorted(nums: list) -> bool: 
+            for i in range(1, len(nums)): 
+                if nums[i] <= nums[i - 1]:
+                    return False
+            return True
+        
+        __traverse(root)
+        res = __is_sorted(candidate_list)
+        
+        return res
 
 # 简单递归法
 class Solution: