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Question:
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1 / 2 2 / \ / 3 4 4 3
But the following is not:
1 / 2 2 \ 3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
Analysis:
问题描述：给出一棵二叉树，判断它是否是自己的镜像树。即围绕着中心节点对称。
思路一：遍历整棵二叉树，然后判断每层的节点是否是对称的。
思路二：递归判断。首先判断该节点的左右节点是否对称，然后保存左右节点，依次递归判断左节点的左节点与右节点的右节点是否对称，以及左节点的右节点与右节点的左节点是否对称。
Answer：
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public boolean isSymmetric(TreeNode root) { if(root == null) return true; return judge(root.left, root.right); } public boolean judge(TreeNode left, TreeNode right) { if(left == null && right == null) return true; if(left == null || right == null) return false; return left.val == right.val && judge(left.left, right.right) && judge(left.right, right.left); } }
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原文地址：http://www.cnblogs.com/little-YTMM/p/4854956.html
