从简单的道题目開始刷题目：

Symmetric Tree

 

题目：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

题目分析：

第一道题目简单的题目，主要利用递归方法，保存左右两个结点。对于LeftNode结点和RightNode结点，推断LeftNode的左结点和RightNode的右结点和LeftNode的右结点和RightNode的左结点是否相等就可以，仅仅要有不相等就能够结束。跳出递归。

代码：

/** * 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 check(TreeNode *leftNode,TreeNode *rightNode)    {        if(leftNode == NULL && rightNode == NULL)            return true;        if(leftNode == NULL ||rightNode ==NULL)            return false;        if(leftNode ->val != rightNode->val)            return false;                return  check(leftNode->left,rightNode->right) && check(leftNode->right,rightNode->left);            }        bool isSymmetric(TreeNode *root) {                if(root == NULL)        return true;        return check(root->left,root->right);    }};