Question

I have a list that looks like (A (B (C D)) (E (F))) which represents this tree:

      A     /  \   B      E  / \    / C   D  F 

How do I print it as (A B E C D F) ?

This is as far as I managed:

((lambda(tree) (loop for ele in tree do (print ele))) my-list) 

But it prints:

A (B (C D)) (E (F)) NIL 

I’m pretty new to Common LISP so there may be functions that I should’ve used. If that’s the case then enlight me.

Thanks.

Answer 1

Taking your question at face value, you want to print out the nodes in ‘breadth-first’ order, rather than using one of the standard, depth-first orderings: ‘in-order’ or ‘pre-order’ or ‘post-order’.

• in-order: C B D A E F
• pre-order: A B C D E F

• post-order: C D B F E A

• requested order: A B E C D F

In your tree structure, each element can be either an atom, or a list with one element, or a list with two elements. The first element of a list is always an atom.

What I think the pseudo-code needs to look like is approximately:

Given a list 'remains-of-tree':     Create empty 'next-level' list     Foreach item in `remains-of-tree`         Print the CAR of `remains-of-tree`         If the CDR of `remains-of-tree` is not empty              CONS the first item onto 'next-level'              If there is a second item, CONS that onto `next-level`     Recurse, passing `next-level` as argument. 

I’m 100% sure that can be cleaned up (that looks like trivial tail recursion, all else apart). However, I think it works.

Start: (A (B (C D)) (E (F))) Level 1: Print CAR: A Add (B (C D)) to next-level: ((B (C D))) Add (E (F)) to next-level: ((B (C D)) (E (F))) Pass ((B (C D) (E (F))) to level 2: Level 2: Item 1 is (B (C D)) Print CAR: B Push C to next-level: (C) Push D to next-level: (C D) Item 2 is (E (F)) Print CAR: E Push F to next-level: (C D F) Pass (C D F) to level 3: Level 3: Item 1 is C Print CAR: C Item 2 is D Print CAR: D Item 3 is F Print CAR: F