Python树的子结构

多出需要判断指针是不是None，避免访问空指针而造成程序崩溃。

'''
输入两棵二叉树A，B，判断B是不是A的子结构
空树不是任意一个树的子结构
'''


class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
class Solution:
    def HasSubtree(self, pRoot1, pRoot2):
        result = False
        if pRoot1 != None and pRoot2 != None:
            if pRoot1.val == pRoot2.val:
                result = self.DoesTree1haveTree2(pRoot1, pRoot2)
            if not result:
                result = self.HasSubtree(pRoot1.left, pRoot2)
            if not result:
                result = self.HasSubtree(pRoot1.right, pRoot2)
        return result
    # 用于递归判断树的每个节点是否相同
    # 需要注意的地方是: 前两个if语句不可以颠倒顺序
    # 如果颠倒顺序, 会先判断pRoot1是否为None, 其实这个时候pRoot2的结点已经遍历完成确定相等了, 但是返回了False, 判断错误
    def DoesTree1haveTree2(self, pRoot1, pRoot2):
        if pRoot2 == None:
            return True
        if pRoot1 == None:
            return False
        if pRoot1.val != pRoot2.val:
            return False

        return self.DoesTree1haveTree2(pRoot1.left, pRoot2.left) and self.DoesTree1haveTree2(pRoot1.right, pRoot2.right)

pRoot1 = TreeNode(8)
pRoot2 = TreeNode(8)
pRoot3 = TreeNode(7)
pRoot4 = TreeNode(9)
pRoot5 = TreeNode(2)
pRoot6 = TreeNode(4)
pRoot7 = TreeNode(7)
pRoot1.left = pRoot2
pRoot1.right = pRoot3
pRoot2.left = pRoot4
pRoot2.right = pRoot5
pRoot5.left = pRoot6
pRoot5.right = pRoot7

pRoot8 = TreeNode(8)
pRoot9 = TreeNode(9)
pRoot10 = TreeNode(2)
pRoot8.left = pRoot9
pRoot8.right = pRoot10

S = Solution()
print(S.HasSubtree(pRoot1, pRoot8))