题解 | #二叉#判断t1树中是否有与t2树完全相同的子树#

#coding:utf-8
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param root1 TreeNode类 
# @param root2 TreeNode类 
# @return bool布尔型
#
class Solution:
    def isContains(self , root1 , root2 ):
        # write code here
        ret = False
        #init
        if root1 == None:
            return False
        ret_node = self.compareTrees(root1, root2)
        ret_left = self.isContains(root1.left, root2)
        ret_right = self.isContains(root1.right, root2)
        return ret_node or ret_left or ret_right
    
    def compareTrees(self, root1, root2):
        if root1 == None and root2 == None:
            return True
        if root1 == None or root2 == None:
            return False
        if root1.val != root2.val:
            return False
        ret1 = self.compareTrees(root1.left, root2.left)
        ret2 = self.compareTrees(root1.right, root2.right)
        return ret1 and ret2
#方法二：先序遍历