Description
输入两棵二叉树A,B,判断B是不是A的子结构。(ps:我们约定空树不是任意一个树的子结构)
Solutions
# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def is_subtree(self, pRoot1, pRoot2):
if pRoot2 is None:
return True
if pRoot1 is None:
return False
result = True
if pRoot1.val == pRoot2.val:
result = self.is_subtree(pRoot1.left, pRoot2.left) and self.is_subtree(pRoot1.right, pRoot2.right)
else:
result = False
return result
def HasSubtree(self, pRoot1, pRoot2):
result = False
if pRoot1 is not None and pRoot2 is not None:
if pRoot1.val == pRoot2.val:
result = self.is_subtree(pRoot1, pRoot2)
if not result:
result = self.HasSubtree(pRoot1.left, pRoot2)
if not result:
result = self.HasSubtree(pRoot1.right, pRoot2)
return result
需要注意在 HasSubtree 函数中对 if 的情况讨论。还是要强调对过程的透彻分析,才能写出正确的代码。除了