这题嘛，想到和 isSameTree 很像，没想到拆成两个函数，只想到放一个函数解决。实际递归遍历调 isSameTree 就 ok 了。

Definition for a binary tree node.

class TreeNode(object):

def init(self, x):

self.val = x

self.left = None

self.right = None

class Solution(object):
def isSubtree(self, s, t):
"""
:type s: TreeNode
:type t: TreeNode
:rtype: bool
"""
if not s:
return False
if self.isSameTree(s, t):
return True
return self.isSubtree(s.left, t) or self.isSubtree(s.right, t)

def isSameTree(self, s, t):
    if not t and not s:
        return True
    elif (not t) or (not s) or (s.val != t.val):
        return False
    else:
        return self.isSameTree(s.left, t.left) and self.isSameTree(s.right, t.right)