lk_test.py 1.55 KB
Newer Older
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
"""给定三个字符串 s1、s2、s3,请你帮忙验证 s3 是否是由 s1 和 s2 交错 组成的。

两个字符串 s 和 t 交错 的定义与过程如下,其中每个字符串都会被分割成若干 非空
子字符串


s = s1 + s2 + ... + sn
t = t1 + t2 + ... + tm
|n - m| <= 1
交错 是 s1 + t1 + s2 + t2 + s3 + t3 + ... 或者 t1 + s1 + t2 + s2 + t3 + s3 + ...
注意:a + b 意味着字符串 a 和 b 连接。



示例 1:


输入:s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
输出:true
示例 2:

输入:s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
输出:false
示例 3:

输入:s1 = "", s2 = "", s3 = ""
输出:true


提示:

0 <= s1.length, s2.length <= 100
0 <= s3.length <= 200
s1、s2、和 s3 都由小写英文字母组成"""


# Definition for a binary tree node.
class TreeNode(object):
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right


class Solution(object):
    def isValidBST(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if root is None:
            return True
        if root.left is None and root.right is None:
            return True
        if root.left is None:
            return self.isValidBST(root.right) and root.val < root.right.val
        if root.right is None:
            return self.isValidBST(root.left) and root.val > root.left.val
        return self.isValidBST(root.left) and self.isValidBST(
            root.right) and root.val > root.left.val and root.val < root.right.val