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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
递归解决，代码如下：
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public boolean symmetric(TreeNode l,TreeNode r) { if(l==null&&r==null) return true; if(l==null||r==null) return false; if(l.val!=r.val) return false; if(!(symmetric(l.left,r.right)&&symmetric(l.right,r.left))) return false; return true; } public boolean isSymmetric(TreeNode root) { if(root==null) return true; return symmetric(root.left,root.right); } }
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原文地址：http://www.cnblogs.com/mrpod2g/p/4293053.html
