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.
https://oj.leetcode.com/submissions/detail/7234082/
思路1：递归判断，注意null的处理。
思路2：迭代写法，有空一起整理。
public class Solution { public boolean isSymmetric(TreeNode root) { if (root == null) return true; return check(root.left, root.right); } private boolean check(TreeNode left, TreeNode right) { if (left == null && right == null) return true; if (left == null || right == null) return false; if (left.val != right.val) return false; return check(left.right, right.left) && check(left.left, right.right); } public static void main(String[] args) { TreeNode root = new TreeNode(10); root.left = new TreeNode(5); root.left.right = new TreeNode(9); root.right = new TreeNode(5); root.right.left = new TreeNode(9); System.out.println(new Solution().isSymmetric(root)); } }
参考：
http://www.cnblogs.com/remlostime/archive/2012/11/15/2772230.html
[leetcode] Symmetric Tree,布布扣,bubuko.com
原文：http://www.cnblogs.com/jdflyfly/p/3821304.html
