It's a subtle error. The problem lies in your helper code. Look at the following:

for (int i = 0; i < 27; i++)
{
    if (curs->children[i] != NULL)
    {
        curs = curs->children[i];
        helper(curs);           
    }
}

The reassignment curs = curs->children[i]; is disrupting what you were trying to accomplish. For simplicity, let's assume that there are no nulls in the first 3 levels. On the first pass, when i=0, curs = root, but gets reassigned to root->children[0]. When teh recursion comes back to the top level, instead of processing root[1], it will process root[0]->children[1]. The third pass (i=2) will process root[0]->children[1]->children[2], and so on, until it has nothing left to process.

The point is that when you return from the recursive call, you will not continue to process the same level in the trie, but will be one level further down the trie because of the reassignment. The remainder of the previous level will never get processed. Also, after each call to helper completes the for loop, it will process the final free(curs) and will lose any nodes that were skipped or orphaned.

This is easily fixed by not doing the reassignment. Instead, just call helpers on the actual child. You can also deal with your return issue here. You'll have to add/restore the code to determine a fail, but the return will handle passing the bool true/false back to the top.

`return helpers(curs->children[i]);`

You might also find my answer on recursion at the following link helpful:

Pset3 Binary Search problems

If this answers your question, please click the check mark to accept this and remove the question from the unanswered pool. Let's keep up on forum maintenance. ;-)

It's a subtle error. The problem lies in your helper code. Look at the following:

for (int i = 0; i < 27; i++)
{
    if (curs->children[i] != NULL)
    {
        curs = curs->children[i];
        helper(curs);           
    }
}

The reassignment curs = curs->children[i]; is disrupting what you were trying to accomplish. For simplicity, let's assume that there are no nulls in the first 3 levels. On the first pass, when i=0, curs = root, but gets reassigned to root->children[0]. When teh recursion comes back to the top level, instead of processing root[1], it will process root[0]->children[1]. The third pass (i=2) will process root[0]->children[1]->children[2], and so on, until it has nothing left to process.

The point is that when you return from the recursive call, you will not continue to process the same level in the trie, but will be one level further down the trie because of the reassignment. The remainder of the previous level will never get processed. Also, after each call to helper completes the for loop, it will process the final free(curs) and will lose any nodes that were skipped or orphaned.

This is easily fixed by not doing the reassignment. Instead, just call helpers on the actual child. You can also deal with your return issue here. You'll have to add/restore the code to determine a fail, but the return will handle passing the bool true/false back to the top.

`return helpers(curs->children[i]);`

You might also find my answer on recursion at the following link helpful:

Pset3 Binary Search problems

If this answers your question, please click the check mark to accept this and remove the question from the unanswered pool. Let's keep up on forum maintenance. ;-)