【劍指Offer】18樹的子結構
阿新 • • 發佈:2018-11-11
題目描述
輸入兩棵二叉樹A,B,判斷B是不是A的子結構。(ps:我們約定空樹不是任意一個樹的子結構)
時間限制:1秒;空間限制:32768K
解題思路
解題思路分為兩步,主要用到了遞迴的方法。第一步先遍歷樹A,找到和樹B根節點值相同的節點A';第二步判斷以A'為根節點的子樹中是否包含樹B。
# -*- coding:utf-8 -*- # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: def HasSubtree(self, pRoot1, pRoot2): # write code here result = False if pRoot1!=None and pRoot2!=None: if pRoot1.val == pRoot2.val: result = self.iscomprise(pRoot1,pRoot2) #不寫self.會報錯 if result != True: result = self.HasSubtree(pRoot1.left,pRoot2) if result != True: result = self.HasSubtree(pRoot1.right,pRoot2) return result def iscomprise(self, pRoot1, pRoot2): if pRoot2 == None: return True if pRoot1 == None: return False if pRoot1.val == pRoot2.val: return self.iscomprise(pRoot1.left, pRoot2.left) and self.iscomprise(pRoot1.right, pRoot2.right)