Question

我正在尝试为BTree实现插入方法。我的BTree类包含一个指向根BNode的指针。 insert方法有内部和外部方法。内部的一个还没有递归，但是我在传入root时保持其第二次插入的值时遇到了麻烦。我认为这可能是一个范围问题，但我不确定它究竟是什么。在main中我尝试插入2个值来测试方法，但似乎在第二次插入时它仍然认为root == null并覆盖第一个键值。有谁可能出现这种情况？

以下是我的功能：

    template<typename T, int M>
    class BTree{
    private:


        returnStatus _insert( BNode<T, M>* r, const T& val, BNode<T, M>*& dptr, T& dkv );
        returnStatus shuffle_insert( BNode<T, M>* r, const T& val );
        returnStatus split_insert( BNode<T, M>* r, const T& val, BNode<T, M>*& dptr, T& dkv );
        bool isNodeLeaf( BNode<T, M>* node );
        int _getNodeIndex( BNode<T, M>* r, const T& val );

        returnStatus _find( BNode<T, M>* r, const T& val );
        returnStatus _remove( BNode<T, M>* r, const T& val, BNode<T, M>* & dptr, T& dkv );
        void _traverse( BNode<T, M>* r );

    public:
        BNode<T, M>* root;
        int size;

        BTree();
        void insert( const T& val );
        returnStatus find( const T& val );
        void remove( const T& val );
        void traverse();
        ~BTree();

    };
    template<typename T, int M>
    BTree<T, M>::BTree()
    {
    root = NULL;
    size = 0;
        }
        template<typename T, int M>
    returnStatus BTree<T, M>::_insert( BNode<T, M>* r, const T& val, BNode<T, M>*& dptr, T& dkv )
    {
    if( r == NULL ){
        dptr = NULL;
        dkv = val;
        r = new BNode<T, M>;
        r->keys[0] = val;
        size++; 
        r->keyCount++;

        return unsuccessful;
    }else{
        returnStatus contains = find(val);

        if( contains == duplicate ){
            return duplicate;
        }else{

            if( r->keyCount < M-1 ){
                returnStatus hold = shuffle_insert(r, val);
                return hold;
            }else{
                cout<<"need to split"<<endl;
                //split_insert
            }
        }
    }
}
template<typename T, int M>
returnStatus BTree<T, M>::shuffle_insert( BNode<T, M>* r, const T& val )
{
    if(r->keyCount == M-1 ) return unsuccessful;

    int index = _getNodeIndex(r, val);

    for(int i = r->keyCount; i>index; i--){
        r->keys[i+1] = r->keys[i];
    }
    r->keys[index] = val;
    r->keyCount++;
}
template<typename T, int M>
returnStatus BTree<T, M>::split_insert( BNode<T, M>* r, const T& val, BNode<T, M>*& dptr, T& dkv )
{



}
template<typename T, int M>
bool BTree<T, M>::isNodeLeaf( BNode<T, M>* node )
{
    for( int i = 0; i< M-1; i++ ){
        if( node->ptr[i] != NULL ){
            return false;
        }
    }
    return true;
}
template<typename T, int M>
returnStatus BTree<T, M>::_find( BNode<T, M>* r, const T& val )
{
    for( int i = 0; i<M; i++ ){
        if( r->ptr[i] != NULL ){
            _find( r->ptr[i], val);
        }
    }
    for( int i =0; i<r->keyCount; i++ ){
        if( r->keys[i] == val ) return duplicate;
    }
    return unsuccessful;
}
template<typename T, int M>
int BTree<T, M>::_getNodeIndex( BNode<T, M>* r, const T& val )
{
    for( int i = 0; i < r->keyCount; i++){
        if( val < r->keys[i] ){
            return i;
        }
    }
    return -1;
}

    template<typename T, int M>
void BTree<T, M>::insert( const T& val )
{
    int set = 0;
    BNode<T, M>* myNode = new BNode<T, M>();                        //May not be correct way to instantiate Reference to pointer
    returnStatus status = _insert( root, val, myNode, set);

    if( status == successful ){
        return;
    }else if( status == unsuccessful ){


    }else if( status == duplicate ){
        cout<<val<<" has already been inserted."<<endl;
    }

}
template<typename T, int M>
returnStatus BTree<T, M>::find( const T& val )
{
    returnStatus stat = _find(root, val);
    return stat;
}
template<typename T, int M>
void BTree<T, M>::remove( const T& val )
{

}
template<typename T, int M>
void BTree<T, M>::traverse()
{
    _traverse(root);
}
template<typename T, int M>
BTree<T, M>::~BTree()
{



}

Answer 1

我认为以下插入B-Tree的方法可以帮助您澄清与插入B树相关的疑问。

public void insert (Comparable object)
{
if(isEmpty ())
{
    if(parent==NULL)
    {
        attachSubtree (0, new BTree(getM ()));
        key[1]=object;
        attachSubtree (1, new BTree(getM ()));
        count=1;
    }
    else
        parent.insertPair(object, new BTree (getM ()));
}

else
{
    int index = findIndex (object);
    if(index !=0 && object.isEQ (key [index])) throw new IllegalArgumentException ("duplicate key");
        subtree [index].insert(object);
}
}

这是从书中提取的 "Data Structures and Algorithms with Object-Oriented Design Patterns in Java"

将节点插入B树

1 个答案: