Traversing a single linked list in order

问题

I have been trying to think a way to traverse a single linked list.

This is so far what I have done:

#include <iostream>

typedef struct node {                                                               
      int data;               // will store information
      node *next;             // the reference to the next node
};  


int printList(node *traverse) {
    if (traverse->next == NULL) {
        return -1;
    }
    traverse=traverse->next;
    printList(traverse);
    cout << traverse->data << endl;
    return 0;
}

int main() {
    node *head = NULL;      
    for (int i = 0; i < 10; i++) {
        node *newEntry = new node;
        newEntry->data = i;
        newEntry->next = head;
        head = newEntry;
    }
    printList(head);
    return 0;
}

I cannot think a way to print the last digit(9) in printList() function. How would I be able to achieve this? My second question is, how can I traverse the same in a while loop rather than a recursive function.

As some of you tried to answer before, I am not looking to traverse this from 9 to 0, This should traverse from 0 to 9, you can see the output from http://codepad.org/ynEdGc9S

回答1:

There are a few things here:

---

In main() the way you are creating the list is incorrect. Draw what you're doing, you'll realize that your head is the last item in the list, i.e., it will probably have a value of 9. (Print out head's value just before you call printList to verify this).

Let me explain (follow along in your code) with iteration of i = 1:

Current state: head=[0]

1. A new temporary node is allocated [ ]
2. Then you assign data to it [1]
3. Then you set this temporary node's next to your head [1]-->[0] ; head=[0]
4. Then you set head to this temporary node [1]-->[0] ; head = [1]

So, you can see what's happening here. Head should still be [0] and its next should be [1] not the other way around.

You can explore and think about the correct way of doing this.

---

printList, this is printing out the recursive stack and not the traversal. Traversal would print them in reverse order because your list is in the reverse order (check the previous section ^ for why).

This is the correct way to print the link in a traversal. This will print the elements of the list in the way they are. When you checked for traverse->next==NULL, traverse held the last element. Since you just ended the recursion by returned -1, the last element was never printed.

int printList(node *traverse) {
   if (traverse == NULL) {  
       return -1;
    }
    cout << traverse->data << endl;
    printList(traverse->next);
    return 0;
}

---

Iterative

int printList(node *traverse) {
   while(traverse != NULL) {  
    cout << traverse->data << endl;
    traverse = traverse->next;
   }
}

---

Feel free to post questions, etc.

回答2:

Instead of if (traverse->next == NULL) try if (traverse == NULL)

This way, you print the current node if it's an actual node with data in it. You then recurse. Ultimately, at the end you will recurse into a NULL pointer, which you can easily escape.

回答3:

As an answer to the second part, your code could look like this:

void printList_iter(node* node)
{
    while (node)
    {
        cout << node->data << endl;
        node = node->next;
    }
}

This will loop through the list, printing each element until it gets to a NULL node, which signifies the end of the list. It's a pretty standard iteration algorithm.

回答4:

Why don't you interchange these statements?

traverse=traverse->next;
printList(traverse);
cout << traverse->data << endl;

This should be changed to:

cout << traverse->data << endl;
traverse=traverse->next;
printList(traverse);

This should work. And then, change

if(traverse->next==NULL)

to

if(traverse==NULL)

来源：https://stackoverflow.com/questions/17450085/traversing-a-single-linked-list-in-order