还有BFS的解法，后面补齐

//0027对称二叉树
class Solution {
public:
    //对称的含义就是镜像：左节点的左子树==右节点的右子树
    //                  右节点的左子树==左节点的右子树
    bool Dfs(TreeNode *p, TreeNode *q)
    {
        if (!q || !p) return !q && !p;
        return q -> val == p -> val && Dfs(p -> left, q -> right) && Dfs(p -> right, q -> left);
    }

    bool isSymmetric(TreeNode* root) {
        return !root || Dfs(root -> left, root -> right);
    }
};