Thread: Please help with Binary Tree

The error message says that:
Cannot convert `void *` to `treNode *`

It says the problem is here:
*treePtr = malloc(sizeof(TreeNode));

This is my code:
-------------------------------------

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

struct treeNode
{
struct treeNode *leftPtr;
int data;
struct treeNode *rightPtr;
};

typedef struct treeNode TreeNode;
typedef TreeNode *TreeNodePtr;

void insertNode(TreeNodePtr * ,int);
void inorder(TreeNodePtr);
void preorder(TreeNodePtr);
void postorder(TreeNodePtr);

int main()
{
int i, item;
TreeNodePtr rootPtr =NULL;

srand(time( NULL ));

printf("The #'s being placed in the tree are:\n");

for(i=1;i<=10; i++)
{
item = rand() % 15;
printf("%3d", item);
insertNode( &rootPtr, item );
}
printf("Preorder\n");
preorder( rootPtr );
printf("Inorder\n");
inorder(rootPtr);
printf("Postorder\n");
postorder(rootPtr);
getch();
return 0;
}

void insertNode(TreeNodePtr *treePtr ,int value)
{
if(*treePtr == NULL)
{
*treePtr = malloc(sizeof(TreeNode));

if(*treePtr != NULL )
{
(*treePtr)->data = value;
(*treePtr)->leftPtr = NULL;
(*treePtr)->rightPtr = NULL;
}
else
printf("%d not inserted. No memory available.\n", value);
}
else
if(value < (*treePtr)->data)
insertNode( & ((*treePtr)->leftPtr),value);
else if(value > (*treePtr)->data)
insertNode( & ((*treePtr)->rightPtr),value);
else
printf("dup");
}

void inorder(TreeNodePtr treePtr)
{
if(treePtr != NULL)
{
inorder(treePtr->leftPtr);
printf("%3d",treePtr->data);
inorder(treePtr->rightPtr);
}
}

void preorder(TreeNodePtr treePtr)
{
if(treePtr != NULL)
{
printf("%3d",treePtr->data);
preorder(treePtr->leftPtr);
preorder(treePtr->rightPtr);
}
}

void postorder(TreeNodePtr treePtr)
{
if(treePtr != NULL)
{
postorder(treePtr->leftPtr);
postorder(treePtr->rightPtr);
printf("%3d",treePtr->data);
}
}

You need to compile it as C code rather than C++ code.

malloc returns a void * which means that when you use malloc, you will need to cast the return value to the type of pointer you want.

*treePtr = malloc(sizeof(TreeNode));

Change the above line to:
*treePtr = (treNode *)malloc(sizeof(TreeNode));

>you will need to cast the return value to the type of pointer you want.

No, you don't. Compile it as C.

Sorensen is right, you should compile it as C.

But a different thing, shouldn't

*treePtr = malloc(sizeof(TreeNode));

be

treePtr = malloc(sizeof(TreeNode));

because malloc returns a pointer?

And it seems you forget to free the allocated memory.

But a different thing, shouldn't

*treePtr = malloc(sizeof(TreeNode));

be

treePtr = malloc(sizeof(TreeNode));

because malloc returns a pointer?

*treePtr is a pointer. treePtr is a pointer to a pointer.

No, you don't. Compile it as C.

Some pure C compilers will indentify this as an error just like a C++ compiler will (ie MIPSPro C 7.2).

My view is that the stronger type checking you have the greater portability the code will have.