# -*- 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 if pRoot1 == None or pRoot2 == None: return False else: res0 = self.HasSubtree(pRoot1.left,pRoot2) res1 = self.HasSubtree(pRoot1.right,pRoot2) res2 = self.isSubtree(pRoot1,pRoot2) if res0 == True or res1 == True or res2 == True: return True else: return False def isSubtree(self,pRoot1,pRoot2): if pRoot2 == None: return True if pRoot1 == None: return False if pRoot1.val == pRoot2.val: res0 = self.isSubtree(pRoot1.left,pRoot2.left) res1 = self.isSubtree(pRoot1.right,pRoot2.right) if res0 == True and res1 == True: return True return False
二叉子树判定分为两个步骤,一个是判断是否节点和子树根节点重合,若重合,开始下一步:
第二步,递归判断每个子节点是不是相同,若相同,则返回Ture