Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

   1

  / \

  2 2

 / \ / \

3 4 4 3

But the following [1,2,2,null,3,null,3] is not:

  1

 / \

2  2

 \   \

  3   3

Note:

Bonus points if you could solve it both recursively and iteratively.

思路：

递归法：当要比较的两个结点有一个为空时返回false，当要比较的两个结点值不相等时返回false。 当两结点为空时即为true。

非递归：层次遍历，需要注意的是左结点按照左右的顺序进入队列，而右边结点要按照先右后左顺序进入队列，这样比较的时候才满足是否对称要求。

bool isSymmetric2(TreeNode *root)    {
    //非递归 层次遍历二叉树        TreeNode *left, *right;        if (!root)            return true;        queue
     
       q1, q2;        q1.push(root->left);        q2.push(root->right);        while (!q1.empty() && !q2.empty()){            left = q1.front();            q1.pop();            right = q2.front();            q2.pop();            if (NULL == left && NULL == right)                continue;            if (NULL == left || NULL == right)                return false;            if (left->val != right->val)                return false;            q1.push(left->left);            q1.push(left->right);            q2.push(right->right);            q2.push(right->left);        }        return true;    }    bool isSymmetricCore(TreeNode* left, TreeNode* right)    {        if (left == NULL && right == NULL) return true;        if (left == NULL && right != NULL || left != NULL && right == NULL || left->val != right->val)return false;        return isSymmetricCore(left->left, right->right) && isSymmetricCore(left->right, right->left);//递归判断    }    bool isSymmetric(TreeNode* root)    {        if (root == NULL) return true;        return isSymmetricCore(root->left, root->right);    }
     

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