链接列表和递归的练习题

假定“链表”节点的结构如下。

C

struct Node
{
  int data;
  struct Node *next;
};

Python3

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

C#

public class Node
{ 
    public int data;
    public Node next;
};
 
// This code is contributed by pratham_76

解释以下C函数的功能。
1.以下函数对给定的链表有什么作用?

C++

void fun1(struct Node* head)
{
if(head == NULL)
    return;
 
fun1(head->next);
cout << head->data << " ";
}
 
// This code is contributed by shubhamsingh10

C

void fun1(struct Node* head)
{
  if(head == NULL)
    return;
  
  fun1(head->next);
  printf("%d  ", head->data);
}

Java

static void fun1(Node head)
{
    if (head == null)
    {
        return;
    }
 
    fun1(head.next);
    System.out.print(head.data + " ");
}
 
// This code is contributed by shubhamsingh10

Python

def fun1(head):
    if(head == None):
        return
    fun1(head.next)
    print(head.data, end = " ")
 
# This code is contributed by shubhamsingh10

C＃

static void fun1(Node head)
{
    if (head == null)
    {
        return;
    }
 
    fun1(head.next);
    Console.Write(head.data + " ");
}
 
// This code is contributed by shubhamsingh10

fun1()以相反的方式打印给定的链接列表。对于链接列表1-> 2-> 3-> 4-> 5，fun1()打印5-> 4-> 3-> 2-> 1。
2.以下函数对给定的链表有什么作用?

C++

void fun2(struct Node* head)
{
    if(head == NULL)
        return;
    cout << head->data << " ";
     
    if(head->next != NULL )
        fun2(head->next->next);
    cout << head->data << " ";
}
 
// This code is contributed by shubhamsingh10

C

void fun2(struct Node* head)
{
  if(head == NULL)
    return;
  printf("%d  ", head->data);
 
  if(head->next != NULL )
    fun2(head->next->next);
  printf("%d  ", head->data);  
}

Java

static void fun2(Node head)
{
    if (head == null)
    {
        return;
    }
    System.out.print(head.data + " ");
 
    if (head.next != null)
    {
        fun2(head.next.next);
    }
    System.out.print(head.data + " ");
}
 
// This code is contributed by shubhamsingh10

Python3

def fun2(head):
      
    if(head == None):
        return
    print(head.data, end = " ")
      
    if(head.next != None ):
        fun2(head.next.next)
    print(head.data, end = " ")
 
    # This code is contributed by divyesh072019

C＃

static void fun2(Node head)
{
    if (head == null)
    {
        return;
    }
    Console.Write(head.data + " ");
 
    if (head.next != null)
    {
        fun2(head.next.next);
    }
    Console.Write(head.data + " ");
}
 
// This code is contributed by divyeshrabadiya07

fun2()打印给定链接列表的备用节点，首先从头到尾，然后从头到尾。如果“链表”的节点数为偶数，则fun2()会跳过最后一个节点。对于链表1-> 2-> 3-> 4-> 5，fun2()打印1 3 5 5 3 1.对于链表1-> 2-> 3-> 4-> 5-> 5-> 6，fun2( )打印1 3 5 5 3 1。
下面是一个完整的正在运行的程序，用于测试上述功能。

C++

#include 
using namespace std;
 
/* A linked list node */
class Node
{
    public:
    int data;
    Node *next;
};
 
 
/* Prints a linked list in reverse manner */
void fun1(Node* head)
{
    if(head == NULL)
        return;
     
    fun1(head->next);
    cout << head->data << " ";
}
 
/* prints alternate nodes of a Linked List, first
from head to end, and then from end to head. */
void fun2(Node* start)
{
    if(start == NULL)
        return;
    cout<data<<" ";
     
    if(start->next != NULL )
        fun2(start->next->next);
    cout << start->data << " ";
}
 
/* UTILITY FUNCTIONS TO TEST fun1() and fun2() */
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(Node** head_ref, int new_data)
{
    /* allocate node */
    Node* new_node = new Node();
     
    /* put in the data */
    new_node->data = new_data;
     
    /* link the old list off the new node */
    new_node->next = (*head_ref);
     
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}
 
/* Driver code */
int main()
{
    /* Start with the empty list */
    Node* head = NULL;
     
    /* Using push() to construct below list
        1->2->3->4->5 */
    push(&head, 5);
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);
     
    cout<<"Output of fun1() for list 1->2->3->4->5 \n";
    fun1(head);
     
    cout<<"\nOutput of fun2() for list 1->2->3->4->5 \n";
    fun2(head);
 
    return 0;
}
 
// This code is contributed by rathbhupendra

C

#include
#include
 
/* A linked list node */
struct Node
{
  int data;
  struct Node *next;
};
 
 
/* Prints a linked list in reverse manner */
void fun1(struct Node* head)
{
  if(head == NULL)
    return;
 
  fun1(head->next);
  printf("%d  ", head->data);
}
 
/* prints alternate nodes of a Linked List, first
  from head to end, and then from end to head. */
void fun2(struct Node* start)
{
  if(start == NULL)
    return;
  printf("%d  ", start->data);
 
  if(start->next != NULL )
    fun2(start->next->next);
  printf("%d  ", start->data);
}
 
/* UTILITY FUNCTIONS TO TEST fun1() and fun2() */
/* Given a reference (pointer to pointer) to the head
  of a list and an int, push a new node on the front
  of the list. */
void push(struct Node** head_ref, int new_data)
{
  /* allocate node */
  struct Node* new_node =
          (struct Node*) malloc(sizeof(struct Node));
  
  /* put in the data  */
  new_node->data  = new_data;
  
  /* link the old list off the new node */
  new_node->next = (*head_ref);
  
  /* move the head to point to the new node */
  (*head_ref)    = new_node;
}
  
/* Driver program to test above functions */
int main()
{
  /* Start with the empty list */
  struct Node* head = NULL;
 
  /* Using push() to construct below list
    1->2->3->4->5  */
  push(&head, 5);
  push(&head, 4);
  push(&head, 3);
  push(&head, 2);
  push(&head, 1);  
  
  printf("Output of fun1() for list 1->2->3->4->5 \n");
  fun1(head);
 
  printf("\nOutput of fun2() for list 1->2->3->4->5 \n");
  fun2(head);
         
  getchar();
  return 0;
}

Java

// Java code implementation for above approach
class GFG
{
 
    /* A linked list node */
    static class Node
    {
        int data;
        Node next;
    };
 
    /* Prints a linked list in reverse manner */
    static void fun1(Node head)
    {
        if (head == null)
        {
            return;
        }
 
        fun1(head.next);
        System.out.print(head.data + " ");
    }
 
    /* prints alternate nodes of a Linked List, first
    from head to end, and then from end to head. */
    static void fun2(Node start)
    {
        if (start == null)
        {
            return;
        }
        System.out.print(start.data + " ");
 
        if (start.next != null)
        {
            fun2(start.next.next);
        }
        System.out.print(start.data + " ");
    }
 
    /* UTILITY FUNCTIONS TO TEST fun1() and fun2() */
    /* Given a reference (pointer to pointer) to the head
    of a list and an int, push a new node on the front
    of the list. */
    static Node push(Node head_ref, int new_data)
    {
        /* allocate node */
        Node new_node = new Node();
 
        /* put in the data */
        new_node.data = new_data;
 
        /* link the old list off the new node */
        new_node.next = (head_ref);
 
        /* move the head to point to the new node */
        (head_ref) = new_node;
        return head_ref;
    }
 
    /* Driver code */
    public static void main(String[] args)
    {
        /* Start with the empty list */
        Node head = null;
 
        /* Using push() to conbelow list
        1->2->3->4->5 */
        head = push(head, 5);
        head = push(head, 4);
        head = push(head, 3);
        head = push(head, 2);
        head = push(head, 1);
 
        System.out.print("Output of fun1() for " +
                         "list 1->2->3->4->5 \n");
        fun1(head);
 
        System.out.print("\nOutput of fun2() for " +
                           "list 1->2->3->4->5 \n");
        fun2(head);
    }
}
 
// This code is contributed by Rajput-Ji

Python3

''' A linked list node '''
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
 
''' Prints a linked list in reverse manner '''
def fun1(head):
    if(head == None):
        return
    fun1(head.next)
    print(head.data, end = " ")
 
''' prints alternate nodes of a Linked List, first
from head to end, and then from end to head. '''
def fun2(start):
     
    if(start == None):
        return
    print(start.data, end = " ")
     
    if(start.next != None ):
        fun2(start.next.next)
    print(start.data, end = " ")
 
 
''' UTILITY FUNCTIONS TO TEST fun1() and fun2() '''
''' Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. '''
def push( head, new_data):
     
    ''' put in the data '''
    new_node = Node(new_data)
     
    ''' link the old list off the new node '''
    new_node.next = head
     
    ''' move the head to poto the new node '''
    head = new_node
    return head
 
''' Driver code '''
''' Start with the empty list '''
head = None
 
''' Using push() to construct below list
    1.2.3.4.5 '''
head = Node(5)
head = push(head, 4)
head = push(head, 3)
head = push(head, 2)
head = push(head, 1)
 
print("Output of fun1() for list 1->2->3->4->5")
fun1(head)
 
print("\nOutput of fun2() for list 1->2->3->4->5")
fun2(head)
 
# This code is contributed by SHUBHAMSINGH10

C＃

// C# code implementation for above approach
using System;
 
class GFG
{
 
    /* A linked list node */
    public class Node
    {
        public int data;
        public Node next;
    };
 
    /* Prints a linked list in reverse manner */
    static void fun1(Node head)
    {
        if (head == null)
        {
            return;
        }
 
        fun1(head.next);
        Console.Write(head.data + " ");
    }
 
    /* prints alternate nodes of a Linked List, first
    from head to end, and then from end to head. */
    static void fun2(Node start)
    {
        if (start == null)
        {
            return;
        }
        Console.Write(start.data + " ");
 
        if (start.next != null)
        {
            fun2(start.next.next);
        }
        Console.Write(start.data + " ");
    }
 
    /* UTILITY FUNCTIONS TO TEST fun1() and fun2() */
    /* Given a reference (pointer to pointer) to the head
    of a list and an int,.Push a new node on the front
    of the list. */
    static Node Push(Node head_ref, int new_data)
    {
        /* allocate node */
        Node new_node = new Node();
 
        /* put in the data */
        new_node.data = new_data;
 
        /* link the old list off the new node */
        new_node.next = (head_ref);
 
        /* move the head to point to the new node */
        (head_ref) = new_node;
        return head_ref;
    }
 
    /* Driver code */
    public static void Main(String[] args)
    {
        /* Start with the empty list */
        Node head = null;
 
        /* Using.Push() to conbelow list
        1->2->3->4->5 */
        head = Push(head, 5);
        head = Push(head, 4);
        head = Push(head, 3);
        head = Push(head, 2);
        head = Push(head, 1);
 
        Console.Write("Output of fun1() for " +
                        "list 1->2->3->4->5 \n");
        fun1(head);
 
        Console.Write("\nOutput of fun2() for " +
                        "list 1->2->3->4->5 \n");
        fun2(head);
    }
}
 
// This code is contributed by Rajput-Ji

输出：

Output of fun1() for list 1->2->3->4->5 
5 4 3 2 1 
Output of fun2() for list 1->2->3->4->5 
1 3 5 5 3 1