题目如下:
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] is symmetric: eg: 1 / \ 2 2 / \ / \ 3 4 4 3 true 1 / \ 2 2 \ \ 3 3 false
给定一棵二叉树,让我们来判断它是不是对称的。解这个问题的思路是从根节点出发,分别向左儿子和右儿子两个方向出发遍历二叉树,先看左右儿子节点的值是否相等,不等则返回false。然后调用递归func(left.left, right.right)和func(left.right, right.left),两者均为真时返回true,否则返回false。
代码如下:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public boolean isSymmetric(TreeNode root) { if (root == null) { return true; } return isSymmetric(root.left, root.right); } public boolean isSymmetric(TreeNode left, TreeNode right) { if (left == null || right == null) { return left == right; } if (left.val != right.val) { return false; } return isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left); } }
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