Update 0098.验证二叉搜索树.md

problems/0098.验证二叉搜索树.md

@@ -341,117 +341,113 @@ class Solution {
 
 ## Python 
 
-**递归** - 利用BST中序遍历特性,把树"压缩"成数组
+递归法（版本一）利用中序递增性质，转换成数组
 ```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 isValidBST(self, root: TreeNode) -> bool:
-        # 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]: # ⚠️ 注意: Leetcode定义二叉搜索树中不能有重复元素
-                    return False
-            return True
-        
-        __traverse(root)
-        res = __is_sorted(candidate_list)
-        
-        return res
-```
+    def __init__(self):
+        self.vec = []
 
-**递归** - 标准做法
+    def traversal(self, root):
+        if root is None:
+            return
+        self.traversal(root.left)
+        self.vec.append(root.val)  # 将二叉搜索树转换为有序数组
+        self.traversal(root.right)
 
-```python
-class Solution:
-    def isValidBST(self, root: TreeNode) -> bool:
-        # 规律: BST的中序遍历节点数值是从小到大. 
-        cur_max = -float("INF")
-        def __isValidBST(root: TreeNode) -> bool: 
-            nonlocal cur_max
-            
-            if not root: 
-                return True
-            
-            is_left_valid = __isValidBST(root.left)
-            if cur_max < root.val: 
-                cur_max = root.val
-            else: 
+    def isValidBST(self, root):
+        self.vec = []  # 清空数组
+        self.traversal(root)
+        for i in range(1, len(self.vec)):
+            # 注意要小于等于，搜索树里不能有相同元素
+            if self.vec[i] <= self.vec[i - 1]:
                 return False
-            is_right_valid = __isValidBST(root.right)
-            
-            return is_left_valid and is_right_valid
-        return __isValidBST(root)
+        return True
+
 ```
-**递归** - 避免初始化最小值做法：
+
+递归法（版本二）设定极小值，进行比较
+
 ```python
 class Solution:
-    def isValidBST(self, root: TreeNode) -> bool:
-        # 规律: BST的中序遍历节点数值是从小到大. 
-        pre = None
-        def __isValidBST(root: TreeNode) -> bool: 
-            nonlocal pre
-            
-            if not root: 
-                return True
-            
-            is_left_valid = __isValidBST(root.left)
-            if pre and pre.val>=root.val: return False
-            pre = root
-            is_right_valid = __isValidBST(root.right)
-            
-            return is_left_valid and is_right_valid
-        return __isValidBST(root)
+    def __init__(self):
+        self.maxVal = float('-inf')  # 因为后台测试数据中有int最小值
+
+    def isValidBST(self, root):
+        if root is None:
+            return True
+
+        left = self.isValidBST(root.left)
+        # 中序遍历，验证遍历的元素是不是从小到大
+        if self.maxVal < root.val:
+            self.maxVal = root.val
+        else:
+            return False
+        right = self.isValidBST(root.right)
+
+        return left and right
+
 ```
+递归法（版本三）直接取该树的最小值
 ```python
-迭代-中序遍历
+# 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 isValidBST(self, root: TreeNode) -> bool:
+    def __init__(self):
+        self.pre = None  # 用来记录前一个节点
+
+    def isValidBST(self, root):
+        if root is None:
+            return True
+
+        left = self.isValidBST(root.left)
+
+        if self.pre is not None and self.pre.val >= root.val:
+            return False
+        self.pre = root  # 记录前一个节点
+
+        right = self.isValidBST(root.right)
+        return left and right
+
+
+
+```
+迭代法
+```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 isValidBST(self, root):
         stack = []
         cur = root
-        pre = None
-        while cur or stack:
-            if cur: # 指针来访问节点，访问到最底层
+        pre = None  # 记录前一个节点
+        while cur is not None or len(stack) > 0:
+            if cur is not None:
                 stack.append(cur)
-                cur = cur.left
-            else: # 逐一处理节点
-                cur = stack.pop()
-                if pre and cur.val <= pre.val: # 比较当前节点和前节点的值的大小
+                cur = cur.left  # 左
+            else:
+                cur = stack.pop()  # 中
+                if pre is not None and cur.val <= pre.val:
                     return False
-                pre = cur 
-                cur = cur.right
+                pre = cur  # 保存前一个访问的结点
+                cur = cur.right  # 右
         return True
 
 ```
-```python
-# 遵循Carl的写法，只添加了节点判断的部分
-class Solution:
-    def isValidBST(self, root: TreeNode) -> bool:
-        # method 2
-        que, pre = [], None
-        while root or que:
-            while root:
-                que.append(root)
-                root = root.left
-            root = que.pop()
-            # 对第一个节点只做记录，对后面的节点进行比较
-            if pre is None:
-                pre = root.val
-            else:
-                if pre >= root.val: return False
-                pre = root.val
-            root = root.right
-        return True
-```
+
 
 ## Go