That part I understand .

it's more how I do it here :

Lets say I have this :

data MessageType = Info
                 | Warning
                 | Error Int
  deriving (Show, Eq)

type TimeStamp = Int

data LogMessage = LogMessage MessageType TimeStamp String
                | Unknown String
  deriving (Show, Eq)

data MessageTree = Leaf
                 | Node MessageTree LogMessage MessageTree
  deriving (Show, Eq)

Now I have to put all the LogMessages in a binary tree.
with this signature : insert :: LogMessage -> MessageTree -> MessageTree

I think I need recursion here but I still not see how I can make the clauses.
Normally you can say somethig like this  [] -> 0  or  1 ->   but here it seems to be all seperate logMessages

Roelof





Sumit Sahrawat, Maths & Computing, IIT (BHU) schreef op 28-2-2015 om 11:53:

Draw the trees, and then try to partition every node into (left, element, right).

For example, in your first example (Node leaf "msg 1" leaf), we get

Node "msg 1"

├── leaf

└── leaf

In the second example,

Node "msg 1"

├── leaf

└── Node "msg 2"     -- A sub-tree

    ├── leaf

    └── leaf

Now, substituting a leaf with a Node in the above tree will lead to larger trees.

For another example, look here: http://en.wikipedia.org/wiki/Tagged_union#Examples

On 28 February 2015 at 14:10, Roelof Wobben <r.wobben@home.nl> wrote:

Hello,

I found out that for 1 item I can do this : node = Node leaf "msg 1" leaf
And for 2 items I can do this : node = Node leaf "msg1"  (Node leaf "msg2" leaf)

But now I wonder how I can make this work with recursion. It seems that I keep the first 2 items and change the 3th one from leaf to another node

Roelof


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--

Regards

Sumit Sahrawat

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