101. Symmetric Tree

problems：
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.
tips:
判断二叉树是否对称，假设有两个节点n1和n2，若树平衡，则n1的左子树的值等于n2右子树的值，n1右子树的值等于n2左子树的值。
solution:

class Solution {
public:
    bool isSymmetric(TreeNode* left,TreeNode* right)
{
    if(!left && !right) return true;
    if((!left && right)||(left && !right)||(left->val!=right->val)) return false;
    return isSymmetric(left->left,right->right)&&isSymmetric(left->right,right->left);
}
    bool isSymmetric(TreeNode* root) {
        if(!root) return true;
        return isSymmetric(root->left,root->right);
    }
};