链表遍历有什么问题？

Question

I have to solve a simple problem where i have been provided with the head node of a linked list and i have to return the pointer to the last node.

Here are two implementations:

This one doesn't work(I tried leetcode and geekforgeeks) and it causes a SEGMENTATION FAULT:

node* traversal(node* head){
    node* temp=head;
    while(temp!=NULL) temp=temp->next;

    return temp;
}

This one works just fine:

node* traversal(node* head){
    node *tail,*temp=head;

    while(temp!=NULL){
        tail=temp;
        temp=temp->next;
    }

    return tail;
}

kindly tell me what's the fault in the first code because according to me both the codes are identical....however the 1st always gives SEGMENTATION FAULT

Answer 1

The issue in the first code chunk is : The moment loop will break temp is pointing to NULL and the same pointer is getting returned ie pointer to the last node.

There needs to take one reference pointer which store the last node reference when the loop iterator ie temp move to the next of the last node meaning temp is having NULL value then the function is getting returned.

In order to resolve this two variable will be required in order to get the last node of the link list:

node *tail; // to assign the reference 
node *next = head; // to iterate through the link list.

Answer 2

The condition being checked in the first case needs modification

node* traversal(node* head){
   node* temp=head;
   //when next node is null, that means this is the last node
   //break out of the while loop when you are at last node
   while(temp->next !=NULL){ 
      temp=temp->next;
   }
   //Return the pointer to the last node
   return temp;
}

As per your original code, you would have temp pointing to null and this when you try to dereference from the calling function will return SEGMENTATION FAULT .