Question

#include <iostream>

using namespace std;

#define YES 1
#define NO 0

class tree
{
    private:
        struct leaf
        {
            int data;
            leaf *l;
            leaf *r;
        };
        struct leaf *p;

    public:
        tree();
        ~tree();
        void destruct(leaf *q);
        tree(tree& a);
        void findparent(int n,int &found,leaf* &parent);
        void findfordel(int n,int &found,leaf *&parent,leaf* &x);
        void add(int n);
        void transverse();
        void in(leaf *q);
        void pre(leaf *q);
        void post(leaf *q);
        void del(int n);
        leaf*&  createBST(int *preOrder, int* inOrder, int len);

};

tree::tree()
{
    p=NULL;
}

tree::~tree()
{
    destruct(p);
}

void tree::destruct(leaf *q)
{
    if(q!=NULL)
    {
        destruct(q->l);
        del(q->data);
        destruct(q->r);
    }
}
void tree::findparent(int n,int &found,leaf *&parent)
{
    leaf *q;
    found=NO;
    parent=NULL;

    if(p==NULL)
        return;

    q=p;
    while(q!=NULL)
    {
        if(q->data==n)
        {
            found=YES;
            return;
        }
        if(q->data>n)
        {
            parent=q;
            q=q->l;
        }
        else
        {
            parent=q;
            q=q->r;
        }
    }
}

void tree::add(int n)
{
    int found;
    leaf *t,*parent;
    findparent(n,found,parent);
    if(found==YES)
        cout<<"\nSuch a Node Exists";
    else
    {
        t=new leaf;
        t->data=n;
        t->l=NULL;
        t->r=NULL;

        if(parent==NULL)
            p=t;
        else
            parent->data > n ? parent->l=t : parent->r=t;
    }
}

void tree::transverse()
{
    int c;
    cout<<"\n1.InOrder\n2.Preorder\n3.Postorder\nChoice: ";
    cin>>c;
    switch(c)
    {
        case 1:
            in(p);
            break;

        case 2:
            pre(p);
            break;

        case 3:
            post(p);
            break;
    }
}

void tree::in(leaf *q)
{
    if(q!=NULL)
    {
        in(q->l);
        cout<<"\t"<<q->data<<endl;
        in(q->r);
    }

}

void tree::pre(leaf *q)
{
    if(q!=NULL)
    {
        cout<<"\t"<<q->data<<endl;
        pre(q->l);
        pre(q->r);
    }

}

void tree::post(leaf *q)
{
    if(q!=NULL)
    {
        post(q->l);
        post(q->r);
        cout<<"\t"<<q->data<<endl;
    }

}

void tree::findfordel(int n,int &found,leaf *&parent,leaf *&x)
{
    leaf *q;
    found=0;
    parent=NULL;
    if(p==NULL)
        return;

    q=p;
    while(q!=NULL)
    {
        if(q->data==n)
        {
            found=1;
            x=q;
            return;
        }
        if(q->data>n)
        {
            parent=q;
            q=q->l;
        }
        else
        {
            parent=q;
            q=q->r;
        }
    }
}

void tree::del(int num)
{
    leaf *parent,*x,*xsucc;
    int found;

    // If EMPTY TREE
    if(p==NULL)
    {
        cout<<"\nTree is Empty";
        return;
    }
    parent=x=NULL;
    findfordel(num,found,parent,x);
    if(found==0)
    {
        cout<<"\nNode to be deleted NOT FOUND";
        return;
    }

    // If the node to be deleted has 2 leaves
    if(x->l != NULL && x->r != NULL)
    {
        parent=x;
        xsucc=x->r;

        while(xsucc->l != NULL)
        {
            parent=xsucc;
            xsucc=xsucc->l;
        }
        x->data=xsucc->data;
        x=xsucc;
    }

    // if the node to be deleted has no child
    if(x->l == NULL && x->r == NULL)
    {
        if(parent->r == x)
            parent->r=NULL;
        else
            parent->l=NULL;

        delete x;
        return;
    }

    // if node has only right leaf
    if(x->l == NULL && x->r != NULL )
    {
        if(parent->l == x)
            parent->l=x->r;
        else
            parent->r=x->r;

        delete x;
        return;
    }

    // if node to be deleted has only left child
    if(x->l != NULL && x->r == NULL)
    {
        if(parent->l == x)
            parent->l=x->l;
        else
            parent->r=x->l;

        delete x;
        return;
    }
}


leaf*& tree::createBST(int *preOrder, int* inOrder, int len)
{
    int i;
    bst = new leaf;
//  tree bst;
    if(len < 0)
        return bst;

    bst->data = *preOrder;
    for(i = 0; i < len; i++)
        if(*(inOrder + i) == *preOrder)
        break;
    bst->l = createBST(preOrder + 1, inOrder, i);
    bst->r = createBST(preOrder + i +1, inOrder + i + 1, len-i-1);
    return bst;

}

int main()
{
/*  tree t;
    int data[]={32,16,34,1,87,13,7,18,14,19,23,24,41,5,53};
    for (int iter=0; iter<15; iter++)
    {
        t.add(data[iter]);
    }
    t.transverse();
    t.del(16);
    t.transverse();
    tdel(41);
    t.tranverse();
*/  

    tree bst;
    int pre_data[] = {20,8,4,12,10,14,22};
    int in_data[] = {4,8,10,12,14,20,22};
    bst.createBST(pre_data, in_data, 7);
    bst.transverse();

    return 0;
}

我收到了这样的错误：

mybst.cpp：262：错误：在'*'标记之前预期的构造函数，析构函数或类型转换

主要原因在于此功能：

叶*＆安培; tree :: createBST（int preOrder， int inOrder，int len）

此算法由how to rebuild BST using {pre,in,post}order traversals results

提供

为什么我收到此错误？

Answer 1

改变这个：

leaf*& tree::createBST(int *preOrder, int* inOrder, int len)

到此：

tree::leaf*& tree::createBST(int *preOrder, int* inOrder, int len)

祝你好运。

Answer 2

使用@ybungalobill更正，你也应该这样做以避免语法错误 -

tree::leaf*& tree::createBST(int *preOrder, int* inOrder, int len)
{
    tree::leaf *bst = new tree::leaf;
    // ...

}

从preorder数组和inorder数组构造bst时出错

2 个答案: