[Python手撕]判斷二叉搜尋樹

# 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: Optional[TreeNode]) -> bool:

        def dfs(root):
            if not root:
                return True
            else: 
                left = dfs(root.left)
                if root.val > self.last:
                    self.last = root.val
                else:
                    return False
                right = dfs(root.right)
            return left and right

        self.last = -float('inf')
        return dfs(root)