Returning a structure in C

The need:

     This program is basically solution of a problem I found on internet. Problem statement is:
There is a purse filled with some coins of gold, silver ans copper. A person draws coins one by one and notes down the type of coin each time. He writes g for gold, s for silver and c for copper coin. Thus a character string is formed. You have to write a function countCoins() which takes that string as parameter and returns count each type of coin. As we know that a function can return only single value, we have 2 options left to return all three counts with one function:
    1. store 3 answers in some contiguous location and return a pointer to that location.
    2. If possible, club all three variables to make a single variable and return that single variable. Yes it is possible, we can make a structure variable containing all three counts and return that structure variable.

The code:

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#include<stdio.h>
typedef struct
{
    int silver;
    int copper;
    int gold;
}CoinPurse;
CoinPurse countCoin(char *);

int main()
{
    int i;
    CoinPurse a,*b;
    char *coins="gscgcc";
    a=countCoin(coins);
    printf("Coppper coins are %d\n",a.copper);
    printf("Gold coins are %d\n",a.gold);
    printf("Silver coins are %d\n",a.silver);
    return 0;
}

CoinPurse countCoin(char *coins)
{
    CoinPurse x;
    x.copper=0;
    x.gold=0;
    x.silver=0;
    while(*coins)
    {
        switch(*coins)
        {
        case 'c':    x.copper++;
                    break;
        case 'g':    x.gold++;
                    break;
        case 's':    x.silver++;
                    break;
        default:    break;
        }
        coins++;
    }
    return x;
}

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Approach:
    The approach is very simple. The structure CoinPurse contains counter for each type of coin. the function countCoin() takes a character string , examines each character and based on that increments the correct counter. after the string is finished, the structure containing the counts is returned by the function countCoin(). In this program i have given the character string *coins as "gscgcc" soit is clear that there are 2 gold, 1 silver and 3 copper coins which can be verified by the program output.

Counting total number of nodes in a tree

The need:

     This program counts the total number of nodes in a binary tree and prints that for the user. This count is also known as tree size.

The code:

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#include<stdio.h>
#include<stdlib.h>
struct tree_node
{
 int data;
 struct tree_node *left,*right,*parent;
};
typedef struct tree_node node;
node * new_node(int x);
node * insert_node(node *root, int x);
int count_nodes(node *root);

int main()
{
 node *root=NULL;
 int i,x,max;
 printf("how many numbers do you want to enter in tree?\n");
 scanf("%d",&max);
 printf("Enter %d numbers \n",max);
 for(i=0;i<max;i++)
 {
  scanf("%d",&x);
  root = insert_node(root, x);
 }
 printf("all numbers inserted into the tree\t press enter to count");
 fflush(stdin);
 getchar();
 printf("\nThere are total %d nodes in this tree\n",count_nodes(root));
    return 0;
}

node * insert_node(node *root, int x)
{
 if(!root)
 {
  root = new_node(x);
  return root;
 }
 if(root->data > x)
  root->left = insert_node(root->left,x);
 else
  root->right = insert_node(root->right,x);
 return root;
}

int count_nodes(node *root)
{
 if(!root)
 return 0;
 else
 return(count_nodes(root->left) + 1 + count_nodes(root->right));
}

node * new_node(int x)
{
 node *furnished;
 furnished = (node*)malloc(sizeof(node));
 furnished->data=x;
 furnished->left=NULL;
 furnished->right=NULL;
 furnished->parent=NULL;
 return furnished;
}

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Approach:     The approach is simple. We have to go by recursion. if the root is null(means there is no tree) then number of nodes are 0. Otherwise if root is not null(the tree exists), the total number of nodes will be
       (number of nodes in left subtree + number of nodes in right subtree + 1 ) 
Here recursive function count_nodes does the job very well.