修改py

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 = []
+
+    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)
+
+    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
+        return True
+
 ```
 
-**递归** - 标准做法
+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. 
-        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: 
-                return False
-            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** - 避免初始化最小值做法：
+递归法（版本三）直接取该树的最小值
 ```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:
-        # 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.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
-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 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
 

problems/0108.将有序数组转换为二叉搜索树.md

@@ -316,73 +316,65 @@ class Solution {
 ```
 
 ## Python  
-**递归**
-
+递归法
 ```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 sortedArrayToBST(self, nums: List[int]) -> TreeNode:
-        '''
-        构造二叉树：重点是选取数组最中间元素为分割点，左侧是递归左区间；右侧是递归右区间
-        必然是平衡树
-        左闭右闭区间
-        '''
-        # 返回根节点
-        root = self.traversal(nums, 0, len(nums)-1)
-        return root
-
     def traversal(self, nums: List[int], left: int, right: int) -> TreeNode:
-        # Base Case
         if left > right:
             return None
 
-        # 确定左右界的中心，防越界
         mid = left + (right - left) // 2
-        # 构建根节点
-        mid_root = TreeNode(nums[mid])
-        # 构建以左右界的中心为分割点的左右子树
-        mid_root.left = self.traversal(nums, left, mid-1)
-        mid_root.right = self.traversal(nums, mid+1, right)
-
-        # 返回由被传入的左右界定义的某子树的根节点
-        return mid_root
+        root = TreeNode(nums[mid])
+        root.left = self.traversal(nums, left, mid - 1)
+        root.right = self.traversal(nums, mid + 1, right)
+        return root
+
+    def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
+        root = self.traversal(nums, 0, len(nums) - 1)
+        return root
+
 ```
 
-**迭代**（左闭右开）
+迭代法
 ```python
+from collections import deque
+
 class Solution:
-    def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:
-        if len(nums) == 0: return None
-        root = TreeNode()  # 初始化
-        nodeSt = [root]
-        leftSt = [0]
-        rightSt = [len(nums)]
-
-        while nodeSt:
-            node = nodeSt.pop()  # 处理根节点
-            left = leftSt.pop()
-            right = rightSt.pop()
-            mid = left + (right - left) // 2
-            node.val = nums[mid]
-
-            if left < mid:  # 处理左区间
-                node.left = TreeNode()
-                nodeSt.append(node.left)
-                leftSt.append(left)
-                rightSt.append(mid)
-
-            if right > mid + 1:  # 处理右区间
-                node.right = TreeNode()
-                nodeSt.append(node.right)
-                leftSt.append(mid + 1)
-                rightSt.append(right)
+    def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
+        if len(nums) == 0:
+            return None
 
+        root = TreeNode(0)  # 初始根节点
+        nodeQue = deque()   # 放遍历的节点
+        leftQue = deque()   # 保存左区间下标
+        rightQue = deque()  # 保存右区间下标
+
+        nodeQue.append(root)               # 根节点入队列
+        leftQue.append(0)                  # 0为左区间下标初始位置
+        rightQue.append(len(nums) - 1)     # len(nums) - 1为右区间下标初始位置
+
+        while nodeQue:
+            curNode = nodeQue.popleft()
+            left = leftQue.popleft()
+            right = rightQue.popleft()
+            mid = left + (right - left) // 2
+
+            curNode.val = nums[mid]  # 将mid对应的元素给中间节点
+
+            if left <= mid - 1:  # 处理左区间
+                curNode.left = TreeNode(0)
+                nodeQue.append(curNode.left)
+                leftQue.append(left)
+                rightQue.append(mid - 1)
+
+            if right >= mid + 1:  # 处理右区间
+                curNode.right = TreeNode(0)
+                nodeQue.append(curNode.right)
+                leftQue.append(mid + 1)
+                rightQue.append(right)
+
         return root
+
 ```
 
 ## Go 

problems/0110.平衡二叉树.md

@@ -536,18 +536,16 @@ class Solution:
 
 ```python
 class Solution:
-    def isBalanced(self, root: TreeNode) -> bool:
-        return self.height(root) != -1
-    def height(self, node: TreeNode) -> int:
-            if not node:
-                return 0
-            left = self.height(node.left)
-            if left == -1:
-                return -1
-            right = self.height(node.right)
-            if right == -1 or abs(left - right) > 1:
-                return -1
-            return max(left, right) + 1
+    def isBalanced(self, root: Optional[TreeNode]) -> bool:
+        return self.get_hight(root) != -1
+    def get_hight(self, node):
+        if not node:
+            return 0
+        left = self.get_hight(node.left)
+        right = self.get_hight(node.right)
+        if left == -1 or right == -1 or abs(left - right) > 1:
+            return -1
+        return max(left, right) + 1
 ```