simple abstract syntax tree and eval

node.c

#include <stdio.h>
#include <stdlib.h>
typedef union{
    int i;
    char c;
}_value;

typedef struct _node{
    struct _node* left;
    struct _node* right;
    _value value;
}node;

int eval(node* current){
    if(current->value.c == '+') return eval(current->left) + eval(current->right);
    if(current->value.c == '-') return eval(current->left) - eval(current->right);
    if(current->value.c == '/') return eval(current->left) / eval(current->right);
    if(current->value.c == '*') return eval(current->left) * eval(current->right);

    else return current->value.i;
}

void visit(_value v){
    if(v.i > 9) printf("%c", v.c);
    else printf("%d", v.i);
}

void preorder(node* current){
    if(current != NULL){
        visit(current->value);
        preorder(current->left);
        preorder(current->right);
    }
}

void inorder(node* current){
    if(current != NULL){
        inorder(current->left);
        visit(current->value);
        inorder(current->right);

    }
}

void postorder(node* current){
    if(current != NULL){
        postorder(current->left);
        postorder(current->right);
        visit(current->value);
    }
}

int main(int argc, const char *argv[]){
    size_t node_numbers = 7;
    size_t i;

    node** nodes = (node**)malloc(sizeof(node)*node_numbers);
    for(i=0;i<node_numbers;i++) nodes[i] = (node*)malloc(sizeof(node));

    nodes[0]->value.c = '+'; // root node
    nodes[1]->value.c = '-';
    nodes[2]->value.i = 4;
    nodes[3]->value.i = 1;
    nodes[4]->value.c = '*';
    nodes[5]->value.i = 2;
    nodes[6]->value.i = 3;

    nodes[0]->left = nodes[1];
    nodes[0]->right = nodes[2];

    nodes[1]->left = nodes[3];
    nodes[1]->right = nodes[4];

    nodes[4]->left = nodes[5];
    nodes[4]->right = nodes[6];

    printf("%d\n", eval(nodes[0]));
    preorder(nodes[0]);
    printf("\n");
    inorder(nodes[0]);
    printf("\n");
    postorder(nodes[0]);

    return 0;
}