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Symmetric Tree: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.
题意：判断二叉树是否对称。
思路：使用递归的方法同时对两棵子树进行递归判断。
代码：
public class Solution { public boolean isSymmetric(TreeNode root) { if(root==null) return true; return ifSymmetric(root.left,root.right); } public boolean ifSymmetric(TreeNode tree1,TreeNode tree2){ if(tree1==null&&tree2==null) return true; else if(tree1==null || tree2==null) return false; if(tree1.val !=tree2.val) return false; else return (ifSymmetric(tree1.left,tree2.right) && ifSymmetric(tree1.right,tree2.left)); } }
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原文地址：http://www.cnblogs.com/Lewisr/p/5232165.html
