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[leetcode] Symmetric Tree

时间:2014-07-03 19:20:42      阅读:189      评论:0      收藏:0      [点我收藏+]

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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

 

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。

bubuko.com,布布扣
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));

    }

}
View Code

 

参考:

http://www.cnblogs.com/remlostime/archive/2012/11/15/2772230.html

 

[leetcode] Symmetric Tree,布布扣,bubuko.com

[leetcode] Symmetric Tree

标签:style   blog   http   color   strong   os   

原文地址:http://www.cnblogs.com/jdflyfly/p/3821304.html

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