August 26, 2024
P65 - Layout a binary tree (2).
An alternative layout method is depicted in the illustration opposite. Find out the rules and write the corresponding method. Hint: On a given level, the horizontal distance between neighboring nodes is constant.
Use the same conventions as in problem P64.
> Node('a', Node('b', End, Node('c')), Node('d')).layoutBinaryTree2
res0: PositionedNode[Char] = T[3,1]('a T[1,2]('b . T[2,3]('c . .)) T[5,2]('d . .))The tree at right may be constructed with Tree.fromList(List('n','k','m','c','a','e','d','g','u','p','q')). Use it to check your code.
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