Question

我在C ++中的二叉树中遇到插入函数问题。正确插入节点，直到我需要再次向右或向左添加节点。该函数认为我在左侧或右侧没有任何节点，因为我已经在那些地方插入了节点。

这是我的代码：

void insert(string data)
{    
    srand(time(NULL));
    int r;
    node *aux=head;
    node *n=new node(data);
    if (head==NULL)
    {
        head =n;
        return;
    }

    while (aux!=NULL)
    { 
        r=rand()%100;
        if (r>50)
        {
            cout<<"\nRandom is "<<r<<", Therefore we have to go to the right."<<endl;
            aux=aux->right;  
        }
        else
        {   
            cout<<"\nRandom is "<<r<<", Therefore we have to go to the left."<<endl;
            aux=aux->left;
            if (aux!=NULL)
            {
                cout<<aux->getdata()<<endl;
            }
        }
    }

    aux=n;
    cout<<"\nWe insert "<<aux->getdata()<<endl;
}

Answer 1

以下是对代码的轻微修改：

void insert(string data)
      {    srand(time(NULL));
           int r;
           node *aux=head;
           node *n=new node(data);
           if(head==NULL){
                          head =n;
                          return;
                          }

       while(aux!=NULL) // We could put while(true) here.
       { 
                       r=rand(); // Modulo is a somehow slow operation
                       if((r  & 1 )== 0) // This is much faster. It checks if r is even
                       {  cout<<"\nRandom is "<<r<<", which is even therefore we have to go to the right."<<endl;
                          if ( aux->right == NULL) // We found an empty spot, use it and break
                          {
                              aux->right = n; break;
                          }
                          else // else move to the right child and continue
                          {
                              aux=aux->right;  
                              cout<<aux->getdata()<<endl;
                          }
                       }
                       else
                       {   
                           cout<<"\nRandom is "<<r<<", which is odd Therefore we have to go to the left."<<endl;
                          if ( aux->left == NULL) // We found an empty spot, use it and break
                          {
                              aux->left = n; break;
                          }
                          else // else move to the left child and continue
                          {
                              aux=aux->left;  
                              cout<<aux->getdata()<<endl;
                          }

                       }
       }
       cout<<"\nWe insert "<<n->getdata()<<endl;

  }

主要原因是你误用了aux。这是一个例子，我希望能帮助你找出错误：

node * aux = head; // suppose head doesn't have any child node
node * n = new node(data);

aux = aux->left; // Set aux to point on the left child of head
aux = n; // Set aux to point on n

cout << aux == NULL?"Aux is null":"Aux is not null" << endl;
cout << head->left == NULL?"Left is null":"Left is not null" << endl;

这段代码回复：

Aux is not null
Left is null

原因是，当我们将 n 分配给辅助时，我们只是告诉辅助指向 n 而不是指向左侧节点。我们没有指定 n 作为头部的左孩子。

您也可以通过将aux声明为节点指针的指针来解决此问题。

node * * aux = &head;

插入函数在随机二叉树中

1 个答案: