Subtree of Another Tree
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题目详情:https://leetcode.com/problems/subtree-of-another-tree/description/
暴力破解,依次检查以s中的各个节点为根节点的树是否和t相等
class Solution(object): def isSubtree(self, s, t): """ :type s: TreeNode :type t: TreeNode :rtype: bool """ if s and t: return self.compareTrees(s,t) or self.isSubtree(s.left,t) or self.isSubtree(s.right,t) else: #一定不会同时为空的,因为题目给出的条件就是两个不为空的树 return False def compareTrees(self,root1,root2):#比较以root1为根节点的树,以root2为根节点的树是否相同 if root1 and root2:#root1和root2都不为空 if root1.val==root2.val:#如果root1.val和root2.val相等,那么继续比较它们的左右子树 return self.compareTrees(root1.left,root2.left) and self.compareTrees(root1.right,root2.right) else:#root1.val和root2.val不相等,那么返回假值 return False elif root1==None and root2==None:#root1和root2都为空 return True else:#一个为空,另一个不为空 return False
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