Question

我正在研究二叉树程序，当调用我的插入函数时，它似乎插入了输入的值，当检查该值时，我什么都没得到，现在我很难过，我可能会遗漏什么或可能是什么错误？我已经提供了类数据和下面的插入/搜索功能定义，我只是有一种奇怪的感觉，这与我的模板实现或一般帮助中的模板有关吗？

struct node
{
 string keyValue; //value to stored that is being searched for in our tree
 node *left; //left side of pointing tree from parent node
 node *right; //right side of pointing tree from parent node

};
template <class T> 
class Tree
{
  public: 
   Tree(); //constructor
  ~Tree(); //destructor

void displayTree();     
void insertTree(T key);
node *searchTree(T key);
void deleteTree();

private:

 node *root; //points to the root point of our tree
 void displayTree(node *leaf);
 void insertTree(T key, node *leaf); //takes in our keyValue to 
 void deleteTree(node*leaf);
 node *search(T key, node *leaf);
};

 template <class T>
 void Tree<T>::insertTree(T key)
 {
  cout << "instert1" << endl;
   if(root != NULL)
   {
    cout << "instert2" << endl;
    insertTree(key, root);
   }
   else
   {
    cout << "inster else..." << endl;
    root = new node;
    root->keyValue = key;
    cout << root->keyValue << "--root key value" << endl;
    root->left = NULL;
    root->right = NULL;
 }   

}

template <class T>      
void Tree<T>::insertTree(T key, node *leaf)
{
 if(key < leaf->keyValue)
 {
   if(leaf->left != NULL)
   { 
    insertTree(key, leaf->left); 
   }
   else //descend tree to find appropriate NULL node to store keyValue (left side of tree)
   {

     leaf->left = new node; //Creating new node to store our keyValue (data)
     leaf->left -> keyValue = key;
     leaf->left -> left = NULL; //Assigning left and right child of current child node to NULL
     leaf->left -> right = NULL;
   }
  }
   else if(key >= leaf->keyValue)
   {
     if(leaf->right != NULL)
     {
      insertTree(key, leaf->right);
     }
     else //descend tree to find appropriate NULL node to store keyValue (right side of tree)
     {
      leaf->right = new node; //Creating new node to store our keyValue (data)
      leaf->right -> keyValue = key;
      leaf->right -> right = NULL; //Assigning left and right child of current child node to NULL
      leaf->right -> left = NULL;
    }
  }     
}
    template <class T>
node *Tree<T>::searchTree(T key)
{
 cout << "searching for...key: " << key << " and given root value:" << endl;
  return search(key, root);
}
template <class T>
node *Tree<T>::search(T key, node*leaf)
{
  if(leaf != NULL)
  {
    cout << "check passed for search!" << endl;
    if(key == leaf->keyValue)
    {
       return leaf;
    }
    if(key < leaf->keyValue)
    {
      return search(key, leaf->left);
    }
    else
    {
      return search(key, leaf->right);
    }

  }

  else 
  {
    cout << key << " Not found...!" << endl;
    return NULL;
  }
}

Answer 1

您的代码似乎有效。你可以在这里查看：http://ideone.com/w9Aa1w

您可能需要做的事情，就是我所做的，是对您的节点结构进行模板化，就像这样......

template <typename T>
struct NodeStruct
{
 T keyValue; //value to stored that is being searched for in our tree
 NodeStruct<T> *left; //left side of pointing tree from parent node
 NodeStruct<T> *right; //right side of pointing tree from parent node
};

然后在你的树类中，将你的新节点类型定义为'node'：

typedef NodeStruct<T> node;

虽然好像插入了。

将来如果您对所遇到的确切问题更加清楚，将会有所帮助，而不是仅发布整个程序，至少确定您遇到问题的代码的子部分。

这个问题太模糊，可能对其他人没有帮助作为参考资料。

二叉树不插入

1 个答案: