[AdamSpeight2008]

Challenge: Binary Tree (ASCII Art)

Challenge: (Medium)

Given a balance parentheses string of a binary tree, produce the equivalent ASCII art version.

Examples

Input: (xx)
Output:

  /\

Input: (x(xx)}
Output:

  /\
   /\

Input: ((xx)(xx))
Output:

   /\
  /\/\

Input: (x(x(xx)))
Output:

  /\
   /\
    /\

This post has been edited by AdamSpeight2008: 31 January 2014 - 06:07 PM

[thava]

seems some thing went wrong while generate the art
(((xx)(xx))((xx)(xx)))
what is the output of the above input

[AdamSpeight2008]

(((xx)(xx))((xx)(xx)))

     /\
    /  \
   /\  /\
  /\/\/\/\

[thava]

any clue,
(x(x(xx)))
and
(((xx)(xx))((xx)(xx)))
both are at same level but different output

[AdamSpeight2008]

thava: Why don't you grab a pen and draw the trees. See what the difference is.

[thava]

well that was a nice clue, i can able to parse it as btree but not able to print it as you expected

Spoiler

    Public Class Btree

        Sub Parse(val As String)
            Dim child As String = val.Substring(1, val.Length - 2)
            Dim Openparam As String = 0, ParamStart As Integer = 0, PaeamEnd As Integer = 0
            If child.Substring(0, 1) = "(" Then
                Openparam = 1
                ParamStart = 1
                While Openparam > 0
                    If child.Substring(ParamStart, 1) = "(" Then
                        Openparam += 1
                    ElseIf child.Substring(ParamStart, 1) = ")" Then
                        Openparam -= 1
                    End If
                    ParamStart += 1
                End While
                LeftLeg = New Btree
                LeftLeg.Parse(child.Substring(Openparam, ParamStart), Me.childLevel + 1)
                LeftLeg.IsLeft = True
            Else
                LeftLeg = Nothing
            End If
            ParamStart = child.IndexOf("(", ParamStart)
            If ParamStart > 0 Then
                Rightleg = New Btree()
                Rightleg.Parse(child.Substring(ParamStart, child.Length - ParamStart), Me.childLevel + 1)
                Rightleg.IsLeft = False
            Else
                Rightleg = Nothing
            End If
        End Sub

        Public Sub parse(val As String, Level As Integer)
            Me.childLevel = Level
            If Btree.TotalLevel < childLevel Then
                Btree.TotalLevel = childLevel
            End If
            Me.Parse(val)
        End Sub

        Public Sub AppendNodes(Lst As List(Of Btree))
            If Lst Is Nothing Then
                Lst = New List(Of Btree)
            End If
            If LeftLeg IsNot Nothing Then
                Lst.Add(Me.LeftLeg)
                Me.LeftLeg.AppendNodes(Lst)
            End If
            If Rightleg IsNot Nothing Then
                Lst.Add(Me.Rightleg)
                Me.Rightleg.AppendNodes(Lst)
            End If
        End Sub

        Public Property childLevel As Integer = 1
        Public Shared Property TotalLevel As Integer = 1
        Public Property LeftLeg As Btree
        Public Property Rightleg As Btree
        Public Property IsLeft As Boolean
        Public ReadOnly Property Artstring As String
            Get
                Dim TREE As String = ""
                If LeftLeg IsNot Nothing Then
                    TREE = "/\"
                Else
                    TREE = "  "
                End If
                If Rightleg IsNot Nothing Then
                    TREE += "/\"
                Else
                    TREE += "  "
                End If
                Return TREE
            End Get
        End Property
    End Class