题解 | #判断是不是二叉搜索树#

误区：单独判断某一个子树是否满足搜索树是不够的

class Solution:
    def isValidBST(self , root: TreeNode) -> bool:
        # write code here
        if not root:
            return True
        if root.left and root.left.val >= root.val:
            return False
        if root.right and root.right.val <= root.val:
            return False
        else:
            return self.isValidBST(root.right) and self.isValidBST(root.left)

解法：如果一个数是搜索树，则它的中序遍历一定是一个递增数列

class Solution:
    def isValidBST(self , root: TreeNode) -> bool:
        # write code here
        # 中序遍历判断是否为单调增序列
        node_list = self.midOrder(root)
        for i in range(len(node_list) - 1):
            if node_list[i] >= node_list[i + 1]:
                return False
        return True
         
    def midOrder(self, node: TreeNode) -> list[int]:
        if not node:
            return []
        return self.midOrder(node.left) + [node.val] + self.midOrder(node.right)