C++ C++；：二叉树删除根节点

我有一个程序，允许用户在二进制搜索树中添加和删除（除其他外）帐户。我已经测试/调试了这段代码，并找到了导致问题的地方，我只是不太确定如何修复它。我可以很好地添加帐户，并且可以删除除根节点以外的所有节点，而不会出错。以下是我的功能：

    if((current->llink==NULL && current->rlink != NULL) || (current->llink != NULL && current->rlink==NULL)){
        if(current->llink==NULL && current->rlink != NULL){
            if(trailcurrent->llink==current){
                trailcurrent->llink=current->rlink;
                delete current;
                current=NULL;
                cout << "Account #:" << acctNum << " has been removed from the list." << endl;
            }
            else{
                trailcurrent->rlink=current->rlink;
                delete current;
                current=NULL;
                cout << "Account #:" << acctNum << " has been removed from the list." << endl;
            }
        }
        else{
            if(trailcurrent->llink==current){
                trailcurrent->llink=current->llink;
                delete current;
                current=NULL;
                cout << "Account #:" << acctNum << " has been removed from the list." << endl;
            }
            else{
                trailcurrent->rlink=current->llink;
                delete current;
                current=NULL;
                cout << "Account #:" << acctNum << " has been removed from the list." << endl;
            }
        }
        return;
    }
    if(current->llink==NULL && current->rlink==NULL){
        if(trailcurrent->llink==current) // This 'if' statement crashes the program with a read violation when deleting the root.
            trailcurrent->llink=NULL;
        else
            trailcurrent->rlink=NULL;
        delete current;
        cout << "Account #:" << acctNum << " has been removed from the list." << endl;
        return;
    }
    if(current->llink != NULL && current->rlink != NULL){
        nodeType<accountType>* check=current->rlink;
        if((current->llink==NULL)&&(current->rlink==NULL)){
            current=check;
            delete check;
            current->rlink=NULL;
            cout << "Account #:" << acctNum << " has been removed from the list." << endl;
        }
        else{
            if((current->rlink)->llink!=NULL){
                nodeType<accountType>* leftCurrent;
                nodeType<accountType>* leftTrailCurrent;
                leftTrailCurrent=current->rlink;
                leftCurrent=(current->rlink)->llink;
                while(leftCurrent->llink != NULL){
                    leftTrailCurrent=leftCurrent;
                    leftCurrent=leftCurrent->llink;
                }
                current->info=leftCurrent->info;
                delete leftCurrent;
                leftTrailCurrent->llink=NULL;
                cout << "Account #:" << acctNum << " has been removed from the list." << endl;
            }
            else{
                nodeType<accountType>* temp=current->rlink;
                current->info=temp->info;
                current->rlink=temp->rlink;
                delete temp;
                cout << "Account #:" << acctNum << " has been removed from the list." << endl;
            }
        }
        return;
    }
}

---

我现在要站出来告诉你（a）我们不知道你的数据类型是什么样的；请记住，我们无法看到您的整个项目，只有您可以，并且（b）我可以向您保证两个链接指针的“大于”比较，如

current->rlink>current->llink

，这不仅可能不是您想要做的，而且除非两个指针在逻辑上来自同一个物理单一分配（包括1-pass-end），这甚至不是标准定义的行为，我完全知道你看不到我的代码，也不知道数据类型。作为比较，我的意思是找到一种方法把它们移到树上。（所以在现实中，它将是链接中节点中的数据）我想不出怎么做。你真的需要知道我的数据类型吗？（难道你不能假设它有一个你选择的变量来解释我需要做什么吗？）当删除根节点时，它会被右或左子树（如果它们存在）替换。查看数据结构文本。

if(current->rlink>current->llink){}