Merge pull request #479 from Miraclelucy/master

Update 0098.验证二叉搜索树.md - 增加了python3版本的简单递归法和迭代解法

problems/0098.验证二叉搜索树.md

@@ -343,7 +343,7 @@ Python：
 #         self.val = val
 #         self.left = left
 #         self.right = right
-//递归法
+# 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
@@ -355,6 +355,35 @@ class Solution:
             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
+
+# 简单递归法
+class Solution:
+    def isValidBST(self, root: TreeNode) -> bool:
+        def isBST(root, min_val, max_val):
+            if not root: return True
+            if root.val >= max_val or root.val <= min_val:
+                return False
+            return isBST(root.left, min_val, root.val) and isBST(root.right, root.val, max_val)
+        return isBST(root, float("-inf"), float("inf"))
+
+# 迭代-中序遍历
+class Solution:
+    def isValidBST(self, root: TreeNode) -> bool:
+        stack = []
+        cur = root
+        pre = None
+        while cur or stack:
+            if cur: # 指针来访问节点，访问到最底层
+                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
+                    return False
+                pre = cur 
+                cur = cur.right
+        return True
+
 ```
 Go：
 ```Go

problems/0530.二叉搜索树的最小绝对差.md

@@ -222,6 +222,26 @@ class Solution:
         for i in range(len(res)-1):  // 统计有序数组的最小差值
             r = min(abs(res[i]-res[i+1]),r)
         return r
+        
+# 迭代法-中序遍历
+class Solution:
+    def getMinimumDifference(self, root: TreeNode) -> int:
+        stack = []
+        cur = root
+        pre = None
+        result = float('inf')
+        while cur or stack:
+            if cur: # 指针来访问节点，访问到最底层
+                stack.append(cur)
+                cur = cur.left
+            else: # 逐一处理节点
+                cur = stack.pop()
+                if pre: # 当前节点和前节点的值的差值
+                    result = min(result, cur.val - pre.val)
+                pre = cur
+                cur = cur.right
+        return result
+        
 ```
 Go：
 > 中序遍历，然后计算最小差值