How to "print list from tail to head" recursively？

Question

I'm trying to print each node in a linked list from tail to head using recursion. But why I can't use the highlighting code to realize a recursion?

class Solution {
public:
    vector<int> printListFromTailToHead(ListNode* head) {
        if(!head)
        {
            vector<int> a(0);
            return a;
        }
        else if(!head -> next)
        {
            vector<int> a(1, head -> val);
            return a;
        }
        else
/*
            return printListFromTailToHead(head -> next).push_back(head -> val);
*/
    }
};

Answer 1

I'm trying to print each node in a linked list from tail to head using recursion

Because you provide no info about your list, here I will show you a simple implementation of your desired function(s) using a "std::list". Note that the recursion I provide has no function parameters, no global scope vars, and no static vars (and I use cout, not print).

Also, I use a class form known as a Functor. I recommend them for encapsulating 'small' function(s). Please review literature on functors ... I consider them 'simpler' than the typical class.

#include <iostream>
using std::cout, std::cerr, std::endl, std::flush; // c++17
using std::ostream;

#include <list>
using std::list;

#include <string>
using std::string, std::to_string;



// typedefs are often simpler to read
// user typedefs ---------------vvvvvvvvvvv
typedef list<string>            StrList_t;
typedef list<string>::iterator  StrListIt_t;


// Functor
class F820_Recursive_t
{
   // data attributes of class instance are not global nor static vars
   StrList_t    m_strList;
   StrListIt_t  m_it;

public:
   int operator()() { return exec();  }  // Functor entry

private:

   int exec() // builds and modifies a std::list<string>
      {
         // example: using initializer list syntax
         m_strList = StrList_t { "111", "222", "333", "444", "555", "666", "777", "888" };

         recurseAndReportSize();

         // example: appending to existing list
         for (int i=301; i<309; ++i) {
            string s = "S" + to_string(i);
            m_strList.push_back (s);
         }

         recurseAndReportSize();

         cout << "\n  version cplusplus: " <<  __cplusplus  << "\n\n" << endl;
         return 0;
      }

   void recurseAndReportSize()
      {
         if (0 == m_strList.size()) {
            cout << "\n  empty  list" << flush;
            return;
         }

         cout << "\n\n  recurse over list to count elements ... " << flush;

         m_it = m_strList.begin(); // head to tail
         uint count = recurseCount();

         // report
         cout << count << " elements" << endl;

         // report list contents head to tail
         m_it = m_strList.begin();
         cout << "\n  Head to Tail recursive content report "
              << "\n  [" << coutListFromHeadToTail() << "]" << endl;

         // report list contents  tail to head
         m_it = m_strList.end(); m_it--;
         cout << "\n  Tail to Head recursive content report: "
              << "\n  [" << coutListFromTailToHead() << "]" << endl;
      }

   // recurse with no parameters, no static vars, no global vars
   uint recurseCount( )
      {  //        --^-- no parameters
         if (m_it == m_strList.end()) // RTC (recursion termination clause)
            return 0;

         m_it++; // step to next element

         return (1 + recurseCount()); // tail recursion
      }

   // recurse with no parameters, no static vars, no global vars
   string  coutListFromHeadToTail ( )
      {                        // --^-- no parameters
         cout << *m_it++;
         if (m_it == m_strList.end()) // RTC (recursion termination clause)
            return "";
         cout << ", ";
         return coutListFromHeadToTail(); // tail recursion
      }

   // recurse with no parameters, no static vars, no global vars
   string  coutListFromTailToHead ( )
      {                       // --^-- no parameters
         if (m_it == m_strList.begin()) {
            cout << *m_it;
            return "";
         }
         cout << *m_it << ", ";
         m_it--;
         return coutListFromTailToHead();  // tail recursion
      }

}; // class F820_Recursion_t

int main(int, char**) { return F820_Recursive_t()(); }

Output on my Lubuntu 19.04, g++ v8.3

  recurse over list to count elements ... 8 elements

  Head to Tail recursive content report 
  [111, 222, 333, 444, 555, 666, 777, 888]

  Tail to Head recursive content report: 
  [888, 777, 666, 555, 444, 333, 222, 111]


  recurse over list to count elements ... 16 elements

  Head to Tail recursive content report 
  [111, 222, 333, 444, 555, 666, 777, 888, S301, S302, S303, S304, S305, S306, S307, S308]

  Tail to Head recursive content report: 
  [S308, S307, S306, S305, S304, S303, S302, S301, 888, 777, 666, 555, 444, 333, 222, 111]

  version cplusplus: 201703

Answer 2

If you want to print, then print - don't build a vector.

void printListFromTailToHead(ListNode* head) 
{
    if (head)
    {
        printListFromTailToHead(head->next);
        std::cout << head->val << '\n';
    }
}

If you actually don't want to print anything but produce a vector, you need to rearrange the code a little bit, because push_back does not return anything:

vector<int> reverse_list(ListNode* head) 
{
    if (!head)
    {
        return vector<int>{};
    }
    else // You don't need a special case for a one-element list.
    {
        vector<int> ls = reverse_list(head->next);
        ls.push_back(head->val);
        return ls; 
    }
}

Answer 3

Check out this solution, maybe isn't the better way to do it, but you maybe can take the idea. The thing is, you do not need return an array, also I did not get it why are you returning an vector, you could just call the function again and then print it like the code bellow.

#include <iostream>


using namespace std;


class ListNode {

public:
  ListNode *next;
  int value;

  ListNode() {
    this->next = NULL;
    this->value = 0;
  }

  ListNode(int _v) {
    this->next = NULL;
    this->value = _v;
  }
};

void printListNodeFromTailToHead(ListNode *node) {
  // If the current node is null will
  // exit the recursive function
  if (node == NULL) return ;

  // Else will call the function again with next ListNode
  // and print the current value

  printListNodeFromTailToHead(node->next);
  cout << node->value << endl;
}


int main() {
  ListNode *a = new ListNode(1);
  ListNode *tmp = a;

  for (int i = 2; i < 10; i++) {
    a->next = new ListNode(i);
    a = a->next;
  }

  printListNodeFromTailToHead(tmp);
  return 0;
}

Answer 4

Your function outputs nothing. And it is a bad idea to use a vector to output recursively a list.

The function can look for example the following way as it is shown in a demonstrative program below.

#include <iostream>

struct ListNode
{
    int value;
    ListNode *next;
};

void push_front( ListNode * &head, int value )
{
    head = new ListNode { value, head };
}

std::ostream & printListFromHeadToTail( ListNode * &head, std::ostream &os = std::cout  )
{
    return head == nullptr ? os 
                           : ( os << head->value << ' ', printListFromHeadToTail( head->next, os )  );  
}

std::ostream & printListFromTailToHead( ListNode * &head, std::ostream &os = std::cout  )
{
    return head == nullptr ? os 
                           : ( printListFromTailToHead( head->next, os ), os << head->value << ' ' );  
}

int main() 
{
    const int N = 10;
    ListNode *head = nullptr;

    for ( int i = 0; i < N; i++ ) push_front( head, i );

    printListFromHeadToTail( head ) << '\n';
    printListFromTailToHead( head ) << '\n';

    return 0;
}

Its output is

9 8 7 6 5 4 3 2 1 0 
0 1 2 3 4 5 6 7 8 9