Question

所以我试图编写一个二叉搜索树，这个擦除函数给我一个seg错误。我的Node类有一个临时数据，指向左边和右边的指针。右子节点以及父节点。我有一个函数find（临时密钥），它将节点*返回给保存密钥作为其数据的节点。所以在我的主要内容中，我有

Node<temp>* p = bTree->find(key);

查找功能有效，但是当我这样做时 bTree-＆GT;擦除（P）; 打印我的结果，iit编译，但我得到一个段错误。请帮忙，谢谢。

template <typename temp>
void BinTree<temp>::erase(Node<temp>* n){

  if(!n->lChild && !n->rChild){ //n has no children
    delete n;
    n = nullptr;
  }

  else if(n->lChild && n->rChild){ //n has two children
    Node<temp>* p = n->rChild;
    while(p->lChild)
        p = p->lChild; //p will be n's successor
    n->data = p->data; //set the data to that of the successor
    if(p->rChild){  //take care of p's right child if it has one
        p->parent->lChild = p->rChild;
        p->rChild->parent = p->parent;
    }
    delete p;
    p = nullptr;
  }

  else{ //n only has one child
    if(n->parent){
        if(n->data < n->parent->data){ //n is a left child
            if(n->lChild){ //n has only a left child
                n->lChild->parent = n->parent;
                n->parent->lChild = n->lChild;
            }
            else{ //n has only a right child
                n->rChild->parent = n->parent;
                n->parent->lChild = n->rChild;
            }
        }
        else{ //n is a right child
            if(n->lChild){ //n has only a left child
                n->lChild->parent = n->parent;
                n->parent->rChild = n->lChild;
            }
            else{ //n has only a right child
                n->rChild->parent = n->parent;
                n->parent->rChild = n->rChild;
            }
        }
    }
    else{ //n is  the root and has no parent, but still only has one child
        if(n->lChild)
            root = n->lChild;
        else
            root = n->rChild;
    }
    delete n;
    n = nullptr;
  }
}

Answer 1

你对指针有一些严重的误解。

重要的不仅是您要删除的节点本身，还有指向该节点的其他指针。

例如，看一下erase（）中的第一个案例：

if(!n->lChild && !n->rChild){ //n has no children
  delete n;
  n = nullptr;
}

删除节点n。但是n的父亲仍然存储一个指向n ....的指针。

为什么我得到seg故障（核心转储）呢？

1 个答案: