题解 | #树的子结构#

#     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 pRoot1 TreeNode类 
# @param pRoot2 TreeNode类 
# @return bool布尔型
#
class Solution:
    def Judge(self, pRoot1, pRoot2):
        # 1空2空：true；1空2非空：false；1非空2空：true；
        if not pRoot1 and not pRoot2:
            return True
        if not pRoot1 and pRoot2:
            return False
        if pRoot1 and not pRoot2:
            return True
        if pRoot1.val != pRoot2.val:
            return False
        return self.Judge(pRoot1.left, pRoot2.left) and self.Judge(pRoot1.right, pRoot2.right)
        
    def HasSubtree(self , pRoot1: TreeNode, pRoot2: TreeNode) -> bool:
        # write code here
        if not pRoot1 or not pRoot2: # 若两棵树的根节点有一个为空则为假
            return False
        if self.Judge(pRoot1, pRoot2):  # 以当前两个根节点为根节点的树为真，则真
            return True
        return self.HasSubtree(pRoot1.left, pRoot2) or self.HasSubtree(pRoot1.right, pRoot2)  # pRoot1的根节点的子节点与pRoot2比较