feat: Add python code for problem 94, 98 (azl397985856#215)

* Fix solution on problem 88 with js, fixs azl397985856#62

* Fix solution on problem 88, fixs azl397985856#62

* Python solution for problem 90

* Add python solution for problem 92

* Add python solution for problem 94

* Add python code for problem 98

Author: kant-li

problems/94.binary-tree-inorder-traversal.md

@@ -53,7 +53,7 @@ mid是一个具体的节点，left和right`递归求出即可`
 
 ## 代码
 
-* 语言支持：JS，C++
+* 语言支持：JS，C++，Python3
 
 JavaScript Code：
 
@@ -162,4 +162,46 @@ public:
     }
 };
 ```
+Python Code:
+```Python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def inorderTraversal(self, root: TreeNode) -> List[int]:
+        """
+        1. 递归法可以一行代码完成，无需讨论；
+        2. 迭代法一般需要通过一个栈保存节点顺序，我们这里直接使用列表
+          - 首先，我要按照中序遍历的顺序存入栈，这边用的逆序，方便从尾部开始处理
+          - 在存入栈时加入一个是否需要深化的参数
+          - 在回头取值时，这个参数应该是否，即直接取值
+          - 简单调整顺序，即可实现前序和后序遍历
+        """
+        # 递归法
+        # if root is None:
+        #     return []
+        # return self.inorderTraversal(root.left)\
+        #     + [root.val]\
+        #     + self.inorderTraversal(root.right)
+        # 迭代法
+        result = []
+        stack = [(1, root)]
+        while stack:
+            go_deeper, node = stack.pop()
+            if node is None:
+                continue
+            if go_deeper:
+                # 左右节点还需继续深化，并且入栈是先右后左
+                stack.append((1, node.right))
+                # 节点自身已遍历，回头可以直接取值
+                stack.append((0, node))
+                stack.append((1, node.left))
+            else:
+                result.append(node.val)
+        return result
+```
 

problems/98.validate-binary-search-tree.md

@@ -148,7 +148,7 @@ public:
 };
 ```
 ### 定义法
-* 语言支持：C++
+* 语言支持：C++，Python3
 
 C++ Code：
 ```C++
@@ -208,6 +208,37 @@ public:
     }
 };
 ```
+Python Code:
+```Python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def isValidBST(self, root: TreeNode, area: tuple=(-float('inf'), float('inf'))) -> bool:
+        """思路如上面的分析，用Python表达会非常直白
+        :param root TreeNode 节点
+        :param area tuple 取值区间
+        """
+        if root is None:
+            return True
+        
+        is_valid_left = root.left is None or\
+                   (root.left.val < root.val and area[0] < root.left.val < area[1])
+        is_valid_right = root.right is None or\
+                   (root.right.val > root.val and area[0] < root.right.val < area[1])
+        
+        # 左右节点都符合，说明本节点符合要求
+        is_valid = is_valid_left and is_valid_right
+        
+        # 递归下去
+        return is_valid\
+            and self.isValidBST(root.left, (area[0], root.val))\
+            and self.isValidBST(root.right, (root.val, area[1]))
+```
 ## 相关题目
 
 [230.kth-smallest-element-in-a-bst](./230.kth-smallest-element-in-a-bst.md)