题解 | #判断是不是二叉搜索树#

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
#
# @param root TreeNode类
# @return bool布尔型
#
class Solution:
    def isValidBST(self, root: TreeNode) -> bool:
        # write code here
        if not root:
            return False

        if root.left:
            if root.left.val > root.val:
                return False
        if root.right:
            if root.right.val < root.val:
                return False
        if root.right and root.left:
            return self.dfs(root.left, 'left', root.val) and self.dfs(root.right, 'right', root.val)
        elif root.right:
            return self.dfs(root.right, 'right', root.val)
        elif root.left:
            return self.dfs(root.left, 'left', root.val)
        else:
            return True
    def dfs(self, node: TreeNode, d: str, rootVal: int) -> bool:
        if not node.left and not node.right:
            return True

        if node.left:
            if node.left.val > node.val:
                return False
            if d == 'right':
                if node.left.val < rootVal:
                    return False
        if node.right:
            if node.right.val < node.val:
                return False
            if d == 'left':
                if node.right.val > rootVal:
                    return False
        if node.right and node.left:
            return self.dfs(node.left, d, rootVal) and self.dfs(node.right,d,rootVal)
        elif node.right:
            return self.dfs(node.right, d, rootVal)
        else:
            return self.dfs(node.left, d, rootVal)