Binary Tree segmentation fault - c

Question

Here i have written a code which inserts numbers in Binary tree. But it gives segmentation fault error.
And also it says " note: expected ‘struct tree *’ but argument is of type ‘struct node *’" in line 8.
Here is the code :-
#include<stdio.h>
#include<stdlib.h>
struct tree{
int data;
struct tree *left;
struct tree *right;
};
struct tree* insert(struct tree* node, int data)
{
if(!node){
node=malloc(sizeof(struct tree));
node->data=data;
node->left=node->right=NULL;
return node;
}
else {
if(data>node->data){
node->right= insert(node->right,data);
return node;
}
else{
node->left= insert(node->left,data);
}
return node;
}
}
printtree(struct tree* node)
{
if(node){
printf("%d",node->data);
}
printtree(node->left);
printtree(node->right);
}
main()
{
int i,n;
struct tree *NODE;
NODE= insert(NODE,5);
NODE= insert(NODE,3);
NODE= insert(NODE,8);
printtree(NODE);
}

Answer 1

You use if( node ) but better to use if( node != NULL )
You use if( !node ) but better to use if( node == NULL )
It make code more readable.
You have so many mistakes - so ... I make it in my way (my code formating, etc.).
printtree(node->left); printtree(node->right); was outside if( node != NULL ){} so it try to get NULL->left and NULL->right
Tested - code works.
#include<stdio.h>
#include<stdlib.h>
struct tree{
int data;
struct tree *left;
struct tree *right;
};
struct tree* insert(struct tree* node, int data)
{
if( node == NULL ) {
node = malloc( sizeof(struct tree) );
node->data = data;
node->left = node->right = NULL;
} else {
if( data > node->data ){
node->right = insert(node->right, data);
} else {
node->left = insert(node->left, data);
}
}
return node;
}
void printtree(struct tree* node)
{
if( node != NULL ){
printf("%d\n", node->data);
printtree(node->left);
printtree(node->right);
}
}
int main()
{
struct tree *NODE = NULL;
NODE = insert(NODE, 5);
NODE = insert(NODE, 3);
NODE = insert(NODE, 8);
printtree(NODE);
return 0;
}

Answer 2

The local variable: struct tree* node; is not initialized, so the if (!node) test will have undefined behavior. Unless you assign it something or use it to hold a malloc'd node, the expression in the else block tries to dereference an uninitialized pointer.
You should also get used to idea that a tree can be considered a 'recursive' structure, so any node is a tree, and the top-level tree is simply a node. There's no good reason for two separate types here.

Answer 3

You are still making the error of passing NODE by value. If you want to modify it, you must use a pointer to that pointer.
#include<stdio.h>
#include<stdlib.h>
typedef struct t
{
int data;
struct t *left;
struct t *right;
}tree;
tree* insert(tree **node, int data)
{
if(!(*node))
{
*node=malloc(sizeof(tree));
(*node)->data=data;
(*node)->left=(*node)->right=NULL;
return *node;
}
else
{
if(data>(*node)->data)
{
(*node)->right = insert(&((*node)->right),data);
return *node;
}
else
{
(*node)->left = insert(&((*node)->left),data);
return *node;
}
}
}
void printtree(tree *node)
{
if(node)
{
printf("%d",node->data);
printtree(node->left);
printtree(node->right);
}
}
void freeMemory(tree *node)
{
if(node)
{
freeMemory(node->left);
freeMemory(node->right);
free(node);
}
}
int main()
{
tree *NODE = NULL;
NODE= insert(&NODE,5);
NODE= insert(&NODE,3);
NODE= insert(&NODE,8);
printtree(NODE);
freeMemory(NODE);
return 0;
}
Link: http://ideone.com/OpZWiC