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Thread: An important exam!

  1. #11
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    Quote Originally Posted by Zael View Post
    I don't understand how is it possible to solve this problem with binary tree if it has to be possible that one folder can consist of many subfolders and documents (not only 2, which is the main idea of binary tree)???

    I've been trying to do anything, and my time is running out if I knew anyone here In Zagreb (Croatia) I would already pay and try to understand the code on my own. As far as it goes, i'll get nowhere, and the teacher won't let me go on easy as it seams.. 2 weeks and I still got nowhere, sounds bad as hell. If someone could make this program for me, I would be mostly grateful.

    Anyway thx for enormous help
    You trace through all the objects in a folder by following its child's brothers (follow only brother pointers). The main point of a binary tree is that there are two child pointers. With the structure I gave, you would be using a slightly non-standard interpretation of what left/right child means. Instead you would view it as First-SubFolder-Content / Next-CurrentFolder-Content.
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  3. #12
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    Zael is offline Newbie
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    Ok i unfortunetly give up..

    I have a plan, which includes at least something from the imagined exam. On Friday I am going to hand over a sample of a general tree, no metter of the details, hierarchy, folders and documents..
    It's going to be a simple general tree with > than 2 children. I believe it is a recommended start because I obviously don't understand some theory basics...

    So here is my (and hopefully Our) new quest. I just need a working code with premade functions. I hope someone has what I am looking for, because it's an unavoidable topic for everyone who deals with structures and algorythms. Please post it here, I'll examine, study and retell my menthor things I actually understand.

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    The code I posted, in my opinion, is rather easy to understand. I just need to figure out the memory access violation error that keeps popping up. You might want to take a close look at it.

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