[Leetcode 101]判断对称树 Symmetric Tree
发布时间:2020-12-14 04:15:10 所属栏目:大数据 来源:网络整理
导读:【题目】 Given a binary tree,check whether it is a mirror of itself (ie,symmetric around its center). For example,this binary tree? [1,2,3,4,3] ?is symmetric: 1 / 2 2 / / 3 4 4 3 ? But the following? [1,null,3] ?is not: 1 / 2 2 3 3 【
【题目】 Given a binary tree,check whether it is a mirror of itself (ie,symmetric around its center). For example,this binary tree? 1 / 2 2 / / 3 4 4 3 ? But the following? 1 / 2 2 3 3 【思路】 类似于100判断树是否相同,根节点开始分左右两路比较,三种情况讨论。p和q=null、p或q=null、p和q的val相同迭代。 不同在于mirror正好相反, 对left和right比较,即是fun(p1.left,p2.right)&&fun(p1.right,p2.left)。 【代码】 ? class Solution { public boolean isSymmetric(TreeNode root) { if(root==null) return true; return fun(root.left,root.right); } public boolean fun(TreeNode p1,TreeNode p2) { if(p1==null&&p2==null) return true; if(p1==null||p2==null) return false; if(p1.val==p2.val) return fun(p1.left,p2.right)&&fun(p1.right,p2.left); return false; } } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |