对称的二叉树

实则是 菜鸡一枚。。。先镜像，然后复制，然后比较 两树是否相等。。。代码贴上，去学习大佬的代码了

class Solution {
public:
    bool isSymmetrical(TreeNode* pRoot)
    {
        TreeNode* mirroroot=treeCopy(pRoot);
        mirror(pRoot);
        return doesAandB(pRoot,mirroroot);
    }
    TreeNode *treeCopy(TreeNode* root)   //复制二叉树
    {
        if(root==NULL)
            return NULL;
        TreeNode *c=new TreeNode(root->val);
        c->left=treeCopy(root->left);
        c->right=treeCopy(root->right);
        return c;
    }
    void mirror(TreeNode* pRoot)    //找二叉树的镜像
    {
        if(pRoot==NULL)
            return ;
        if(pRoot->left!=NULL && pRoot->right!=NULL)
        {
            TreeNode *temp=pRoot->left;
            pRoot->left=pRoot->right;
            pRoot->right=temp;
        }
        else if(pRoot->left!=NULL && pRoot->right==NULL)
        {
            pRoot->right=pRoot->left;
            pRoot->left=NULL;
        }
        else if(pRoot->right!=NULL && pRoot->left==NULL)
        {
            pRoot->left=pRoot->right;
            pRoot->right=NULL;
        }
        else
            return ;
        mirror(pRoot->left);
        mirror(pRoot->right);
    }
    bool doesAandB(TreeNode* pRoot,TreeNode* mirrorRoot)  //判断二叉树是否相等
    {
        if(pRoot==NULL && mirrorRoot==NULL)
            return true;
        else if(pRoot==NULL && mirrorRoot!=NULL)
            return false;
        else if(pRoot!=NULL && mirrorRoot==NULL)
            return false;
        if(pRoot->val!=mirrorRoot->val)
            return false;
        return doesAandB(pRoot->left,mirrorRoot->left)&&doesAandB(pRoot->right,mirrorRoot->right);
    }
};