Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node's descendants. The tree scould also be considered as a subtree of itself.

Example 1:
Given tree s:

     3
    / \
   4   5
  / \
 1   2

Given tree t:

   4 
  / \
 1   2

Return true, because t has the same structure and node values with a subtree of s.

Example 2:
Given tree s:

     3
    / \
   4   5
  / \
 1   2
    /
   0

Given tree t:

   4
  / \
 1   2

Return false.

LeetCode：連結

給定兩個非空二叉樹s和t，判斷t是否是s的子樹。s的子樹是指由s中某節點及該節點的所有子節點構成的二叉樹。

特別的，s是其本身的子樹。

# 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
        """
        # 必須都為空才行
        if not s or not t:
            return not s and not t  
        if self.check(s, t):
            return True
        return self.isSubtree(s.left, t) or self.isSubtree(s.right, t)

    def check(self, s, t):
        if not s or not t:
            return not s and not t 
        if s.val != t.val:
            return False
        # 必須左子樹右子樹都得滿足
        return self.check(s.left, t.left) and self.check(s.right, t.right)