题解 | #树的子结构#

# 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 pRoot1 TreeNode类 
# @param pRoot2 TreeNode类 
# @return bool布尔型
#
class Solution:
    # 递归比较
    def diguibijiao(self , pRoot1: TreeNode, pRoot2: TreeNode) -> bool:
        if pRoot1 is None and pRoot2 is not None:
            return False
        if pRoot2 is None or pRoot2 is None:
            return True
        # 判断对应值是否相同，并且左子树和右子树是否相同
        return pRoot1.val == pRoot2.val and self.diguibijiao(pRoot1.left,pRoot2.left) and self.diguibijiao(pRoot1.right,pRoot2.right)

    def HasSubtree(self , pRoot1: TreeNode, pRoot2: TreeNode) -> bool:
        # write code here
        if pRoot2 is None:
            return False
        if pRoot1 is None and pRoot2 is not None:
            return False
        if pRoot1 is None or pRoot2 is None:
            return True
        # 递归比较返回值
        flag1 = self.diguibijiao(pRoot1,pRoot2)
        # 上面比较失败，换根结点比较
        flag2 = self.HasSubtree(pRoot1.left,pRoot2)
        flag3 = self.HasSubtree(pRoot1.right,pRoot2)
        return flag1 or flag2 or flag3