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calling recursion on a very huge set may result int stack overflow...but in the above case this situation should not arise...

To answer second part...if you use recursion, the performance will suffer drastically. Because of so many function calls, stack will be filled fast and even if stack is not overflowed, program will take too much time than iterative method.

arungupta1.blogspot.in/2012/09/print-nn-matrix-in-spiral-order.html

arungupta1.blogspot.in/2012/09/print-nn-matrix-in-spiral-order.html

#include <iostream>

using namespace std;

int main()
{
    int n=5;
   int matrix[5][5]={{1,2,3,4,5},{6,7,8,9,10},{11,12,13,14,15},{16,17,18,19,20},{21,22,23,24,25}};
   for(int i=0;i<5;i++)
   {
      for(int j=0;j<5;j++)
      {
          cout<<matrix[i][j]<<" ";
      }
      cout<<endl;
   }
      
   for(int layer=0;layer<=2;layer++)
   {
       cout<<endl<<"Layer:"<<layer+1<<endl;
       for(int c=layer;c<n-layer;c++)
        cout<<matrix[layer][c]<<" ";
       for(int c=layer+1;c<n-layer;c++)
        cout<<matrix[c][n-layer-1]<<" ";
       for(int c=n-layer-2;c>=layer;c--)
        cout<<matrix[n-layer-1][c]<<" ";
       for(int c=n-layer-2;c>layer;c--)
        cout<<matrix[c][layer]<<" ";
       
   }
   
   return 0;
}

#include <iostream>

using namespace std;

#define n 4

int main()
{
   int matrix[n][n]={{1,2,3,4},{6,7,8,9},{11,12,13,14},{16,17,18,19}};
   
    
   for(int layer=0;layer<=int(n/2);layer++)
   {
      cout<<endl<<"Layer:"<<layer+1<<endl;
       for(int c=layer;c<n-layer;c++)
        cout<<matrix[layer][c]<<" ";
       for(int c=layer+1;c<n-layer;c++)
        cout<<matrix[c][n-layer-1]<<" ";
       for(int c=n-layer-2;c>=layer;c--)
        cout<<matrix[n-layer-1][c]<<" ";
       for(int c=n-layer-2;c>layer;c--)
        cout<<matrix[c][layer]<<" ";
       
   }
   
   return 0;
}

#include <stdio.h>
#define R 3
#define C 6
 
void spiralPrint(int m, int n, int a[R][C])
{
    int i, k = 0, l = 0;
 
    /*  k - starting row index
        m - ending row index
        l - starting column index
        n - ending column index
        i - iterator
    */
 
    while (k < m && l < n)
    {
        /* Print the first row from the remaining rows */
        for (i = l; i < n; ++i)
        {
            printf("%d ", a[k][i]);
        }
        k++;
 
        /* Print the last column from the remaining columns */
        for (i = k; i < m; ++i)
        {
            printf("%d ", a[i][n-1]);
        }
        n--;
 
        /* Print the last row from the remaining rows */
        if ( k < m)
        {
            for (i = n-1; i >= l; --i)
            {
                printf("%d ", a[m-1][i]);
            }
            m--;
        }
 
        /* Print the first column from the remaining columns */
        if (l < n)
        {
            for (i = m-1; i >= k; --i)
            {
                printf("%d ", a[i][l]);
            }
            l++;    
        }        
    }
}

#include<iostream>
#include<stdio.h>
using namespace std;

void print(int size,int r,int c)
{
if(r==0)
printf("%2d ",(size*size)-1-c);
else if(c==size-1)
printf("%2d ",size*(size-1)-r);
else
print(size-1,(size-1-r),(size-2-c));

}
int main()
{
int n;
while(cin>>n)
{
for(int r=0;r<n;r++,cout<<endl)
{
for(int c=0;c<n;c++)
{
print(n,r,c);
}
}
}
return 0;
}