Linked List Tree

I'm having a bit of trouble with making a "tree" representation of a linked list. That is, a root pointer points to two or more other pointers, and they point to 2 or more OTHER pointers, and it branches out from there. I'd prefer to do this over the "train" representation of all the pointers pointing to each other in a line, and having to traverse through the whole list to get to certain 'nodes'.

I just don't know how to make the 'root' pointer point to MORE than one object.

Code:

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#include <iostream>
using namespace std;

struct Animal{
      int Age;
      Animal *next;    //Pointer to the next Animal object
     
      /*Constructor that initially sets ages to 15 and new pointers to 0*/
      Animal():Age(15),next(0){cout<<"Just an animal created.\n";}
      virtual void Act(){};                //Not very important
};
struct Cat: public Animal{
      Cat(){cout<<"Kitty created.\n";}
      void Act(){cout<<"Meow!";};
};
struct Dog: public Animal{
      Dog(){cout<<"Doggy created.\n";}
      void Act(){cout<<"Woof!";};
};

int main(void){
    Animal *root = new Animal;    //Create pointer to new Animal
    root->next = new Dog;        /*Pointer in new Animal goes to a new Dog
(But how can I have this point to BOTH a Dog and a CAT?*/

    cin.get();
    return 0;
}

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I'm not the best with linked lists...I don't really know how to create base pointers that can point to several other pointers. How's this done?

A tree is of this structure:

Code:

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struct Node
{
  int data1;
  char data2;
  string data3;

  Node* left;
  Node* right;
};

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Quote:

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But how can I have this point to BOTH a Dog and a CAT?*/

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;) Have 2 pointers, and point one to a Dog and one to a Cat.

Linked lists are completely different from trees. Binary trees will generally have no inbuilt mechanism by which to move from one branch to another, but can only travel down one branch to a leaf at the end. Linked lists have only one linear path to the end. The thing is, trees aren't very well suited to traversing all elements, and it is in fact relatively difficult to do so; they are much better for searching etc. Linked lists are better for do-this-to-everything-in-the-list operations.

Make the Act() method a pure virtual function by adding =0; to the end rather than brackets.

Write a main() function that asks the user for input, either a dog or a cat. If the user wants a dog, give 'em a dog, and vice versa.

You can do this by declaring a linked list to be full of Animals and then filling with child classes of Animals as you please.

Another question: is there any way to limit the amount of confusing syntax like this?

Code:

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root->next->next->next->next->next->next->next->next->Data;
//or.....
root->right->left->right->left->right->left->right->left->Data;

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Yes. Keep a pointer to the current node you're working with, and traverse the list so that all you have to do is q->Data.

It's inadvisable to have code like you posted, because it assumes you have a certain number of elements in your linked list or tree. If you do not, you'll end up dereferencing a null pointer, causing bad things to happen.