[leetcode] 10. Symmetric Tree

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这次我觉得我的智商太低，想了很久才写出来。题目是让求镜像二叉树判断，题目如下：

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.

OJ‘s Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where ‘#‘ signifies a path terminator where no node exists below.

Here‘s an example:

   1
  /  2   3
    /
   4
         5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

我之前的思路有问题，我是想先做一个函数然后记录到某个子叶的路径，这个路径拿左右来表示，0表示左，1表示右。结果这个函数的递归的版本我没写出来，因为结束弹出方法我实在没找到，然后迭代我觉得实在太烦了。然后就想着换另一种方法：把整个函数的出口另外做一个简单的递归，然后在主函数里面调用一次就行。

题解如下：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool Symmetric(TreeNode *left, TreeNode *right)
    {
        if (left == NULL && right == NULL)
        {
            return true;
        }
        if (left == NULL || right == NULL)
        {
            return false;
        }
        return left->val == right->val && Symmetric(left->left, right->right) && Symmetric(left->right, right->left);
    }

    bool isSymmetric(TreeNode *root)
    {
        return root != NULL ? Symmetric(root->left, root->right) : true;
    }
};

即如上。之后leetcode的题目好像要买它的书之后才能再刷了，不过那是对应的是新出的题目，老的题目还是可以的，先把之前的题目刷完再说吧。

[leetcode] 10. Symmetric Tree

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原文地址：http://www.cnblogs.com/TinyBox/p/4105463.html