深度优先搜索

二叉树创建

class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None


def createTree(l):
    l = l[::-1]
    def create(l):
        a = l.pop()
        if a == '#':
            root = None
        else:
            root = TreeNode(a)
            root.left = create(l)
            root.right = create(l)
        return root
    return create(l)

验证二叉搜索树

思路:刚开始以为直接判断每个节点大于左节点，小于右节点，就可以。测试不通过，原因是隔代有可能出现错误。
后面查阅了一番，大部分的解法思路如下：当前节点值就是左子树的最大值，右子树的最小值，所以只要传递下去即可。
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def validBST(self, root, small, large):
        if root is None:
            return True
        elif root.val <= small or root.val >= large:
            return False
        else:
            return self.validBST(root.left, small, root.val) and self.validBST(root.right, root.val, large)
    def isValidBST(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        return self.validBST(root,-2**32,2**32)