On Thu, 2012-05-24 at 11:07 +0200, tobi wrote:

> (Completely giving up anything I knew about binary trees now, because it 
> wasn't much and it wasn't
> even from a book):
> You said that you can traverse from any node in the tree. 
No, any node can be considered as a root for a subtree.  Anynode.parent
provides what I think your saying.

> Consequently every node has to provide
> such functionality. 
(Aside, I think that separating the functionality from the node is the
key of what I was trying to say. But that is a side issue.)

> I don't deem it necessary to distinguish between root and other nodes. 
Only in that the "tree" happens to be a directed graph that has one
special node.  I told you graphs were interesting.

> The codeis only once there anyway. It could even have helped in your case to 
> "em-parent" (sorry, not a
> native English speaker) a node that you formerly assumed to be the root like
> 
> NewNode.Left = RootNode
> RootNode = NewNode
> 
> As interesting as it sounds, I'm probably not the right person to work on 
> this. I don't have much
> time for (I don't think so but from a pure conscientiousness point I need to 
> prepare my A-level a
> bit) and nobody else wants to wait until I finish the theory ;)
> 
Having been born some time in the middle of the last century, I have no
idea what an A-level is.  I hope it's not painful. :-)

> Regards,
> Tobi

and regards from me too!

But if A-levels are what I think they are, I can recall thinking some 40
years ago, "calculus, as if I am ever going to need that in my life,
ha!".
Wrong.  :-)

Now get back to those books!



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