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

时间复杂度O（n）

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