I understand the "rules" of inserting
void printTree(BTNode* node) {
if (node == NULL)
return;
printTree(node->left);
printf("%c", node->item);
printTree(node->right);
}
in createExp
, the right node may be build with some chars already parsed by the left node. It will happened each time the left node parses more than one char.
To prevent this, createExp
should return an info to where parsing ends. Something like this :
char *createExpTree(BTNode** root, char* prefix)
{
if (*prefix) {
if (!isdigit(*prefix)) {
*root = malloc(sizeof(BTNode));
(*root)->item = *prefix;
(*root)->left = NULL;
(*root)->right = NULL;
}
else {
*root = malloc(sizeof(BTNode));
(*root)->item = *prefix;
(*root)->left = NULL;
(*root)->right = NULL;
return prefix;
}
}
prefix = createExpTree(&(*root)->left, ++prefix);
return createExpTree(&(*root)->right, ++prefix);
}
If you need to keep createExpTree
signature, you can flatten the recursion into an loop like that :
void createExpTree(BTNode** root, char* prefix)
{
BTNode** stack[SIZE];
int stackPosition = 0;
while (*prefix) {
*root = malloc(sizeof(BTNode));
(*root)->left = NULL;
(*root)->right = NULL;
if (isdigit(*prefix)) {
(*root)->item = *prefix++;
if (stackPosition == 0) break;
root = stack[--stackPosition];
}
else {
(*root)->item = *prefix++;
stack[stackPosition++] = &(*root)->right;
root = &(*root)->left;
}
}
}