Hi.
 I do not  why the leftchild and rightchild of the root are null.
why?

#include <stdlib.h>
#include <stdio.h>

struct node_b
{
 int key;
 struct node_b *leftchild;
 struct node_b *rightchild;

typedef struct node_b node;

node *create(int data)
{
 node *root;
 root=malloc(sizeof(node));
 root->key=data;
 root->leftchild=NULL;
 root->rightchild=NULL;
 return root;

void insert(int data, node *b_node)
{
 if(b_node==NULL)
 {
  b_node=malloc(sizeof(node));
  b_node->key=data;
  b_node->leftchild=NULL;
  b_node->rightchild=NULL;
 }
 else
 {
  if(data<b_node->key)
  {
   insert(data, b_node->leftchild);
  }
  else
  {
   insert(data, b_node->rightchild);
  }
 }

void print_node(int low, int high, node *b_node)
{
 if(b_node!=NULL)
 {
  print_node(low, high, b_node->leftchild);
  print_node(low, high, b_node->rightchild);
  print_node(low, high, b_node);
  if(b_node->key<=high&&b_node->key>=low)
  {
   printf(" %d", b_node->key);
  }
 }

void freetree(node *b_node)
{
 if(b_node!=NULL)
 {
  freetree(b_node->leftchild);
  freetree(b_node->rightchild);
  free(b_node);
 }

int main(int argc, char *argv[])
{
 int i=0;
 int indata;
 int numOperand=0;
 node *root_node;
 int low_key=0;
 int up_key=0;

 FILE *infile;
 char DataIn[6];
 strcpy(DataIn, argv[1]);
 infile=fopen(DataIn, "r");

 if(infile==NULL)
 {
  printf("Cannot open file!\n");
  exit(1);
 }

 fscanf(infile, "%d\n", &numOperand);
 for(i=1;i<=numOperand;i++)
 {
  fscanf(infile, "%d", &indata);
  if(i==1)
  {
   root_node=create(indata);
  }
  else
  {
   insert(indata, root_node);
  }
 }
 fscanf(infile, "\n%d", &low_key);
 fscanf(infile, "%d", &up_key);

 printf("The nodes within %d ", low_key);
 printf("to %d ", up_key);
 printf("are \n");
 print_node(low_key, up_key, root_node);

 fclose(infile);

 freetree(root_node);

 return 0;

since you actually allocate b_node (and hence change its value), it must be
passed by reference:

void insert(int data, node **b_node)
{
 if(*b_node==NULL)
 {
  *m_node=malloc(sizeof(node));
 /* ... */

btw please post shorter pieces of code from now on.

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