Question

我正在尝试用c ++创建一个简单的二叉搜索树类，但每当我调用我的插入函数时：

template<typename T>
void BSTree_Fast<T>::insert(T & key)
{
  cout<<"working";
  int i=0;
  BSTNode<T> *newNode=new BSTNode<T>(key);  //Initializing pointer to newNode with key as Key
  if(maxDepth==0){              //if tree is empty, sets root to new node
    root=newNode;
    setMaxDepth(1);
  }
  BSTNode<T> *curr=root;            //Traversal unit curr
  if(key==curr->getKey()){          //checking if inserted key is same as root key
    int n=curr->getNum();
    n++;
    curr->setNum(n);
    delete(newNode);
  }
while(key!=curr->getKey()){         //first while loop when above is not true
  if(key<curr->getKey()){           //checks left if key < current key
    if(curr->getLeft()==NULL){      //if cannot traverse left, creates new node to left of current
        curr->setLeft(newNode);
        if(i>maxDepth){
            cout<<i;
            setMaxDepth(i);
        }
        break;
    }
    else if(curr->getLeft()->getKey()==key){//if left child equals input key, increase number
        int n=curr->getLeft()->getNum();
        n++;
        curr->getLeft()->setNum(n);
        delete(newNode);
        break;
    }
    else{
        i++;
        curr=curr->getLeft();       //traverses left
    }
  }
  if(key>curr->getKey()){           //checks right if key > current key
    if(curr->getRight()==NULL){     //if cannot traverse right, creates new node to right
        curr->setRight(newNode);
        if(i>maxDepth){
            setMaxDepth(i);
        }
        break;
    }
    else if(curr->getRight()->getKey()==key){//if right child equals input key, increase number
        int n=curr->getRight()->getNum();
        n++;
        curr->getRight()->setNum(n);
        delete(newNode);
        break;
    }
    else{
        i++;
        curr=curr->getRight();      //traverses to right
    }
  }
}   
}

然而，在通过从文件读入字符串来调用任何实际代码之前，它似乎遇到了段错误。

Answer 1

完成添加第一个节点的过程：

  BSTNode<T> *newNode=new BSTNode<T>(key);  //Initializing pointer to newNode with key as Key
  if(maxDepth==0){              //if tree is empty, sets root to new node
    root=newNode;
    setMaxDepth(1);
  }

哦，哦，这里没有return。因此，在我们设置root=newNode后，我们继续。

  BSTNode<T> *curr=root;            //Traversal unit curr
  if(key==curr->getKey()){          //checking if inserted key is same as root key
    int n=curr->getNum();
    n++;
    curr->setNum(n);
    delete(newNode);
  }

显然钥匙会匹配。我们刚刚删除了newNode，同时也是root和cur。

while(key!=curr->getKey()){

哎呀，我们访问了刚删除的curr。

c ++ - 函数调用的分段错误？

1 个答案: