#-(and) "

P64 (**) Layout a binary tree (1)

    Given a binary tree as the usual Prolog term t(X,L,R) (or nil). As
    a preparation for drawing the tree, a layout algorithm is required
    to determine the position of each node in a rectangular
    grid. Several layout methods are conceivable, one of them is shown
    in the illustration below.

    [p64]
    In this layout strategy, the position of a node v is obtained by the following two rules:

      □ x(v) is equal to the position of the node v in the inorder sequence
      □ y(v) is equal to the depth of the node v in the tree

    In order to store the position of the nodes, we extend the Prolog
    term representing a node (and its successors) as follows:

    % nil represents the empty tree (as usual)
    % t(W,X,Y,L,R) represents a (non-empty) binary tree with root W
    %  \"positioned\" at (X,Y), and subtrees L and R

    Write a predicate layout-binary-tree/2 with the following specification:

    % layout-binary-tree(T,PT) :- PT is the \"positioned\" binary tree
    % obtained from the binary tree T. (+,?)

    Test your predicate in an appropriate way.
"

(load "p54a")


;; To add the coordinates, we create a new structure, which inherits
;; from the binary-tree structure, so we can reuse that abstraction.
;; However, including structures will make the new fields added at the
;; end of it.  The order of the fields should be immaterial (only that
;; we don't use true structures, but lists, so the new fields are
;; added at the end of the lists, compared to the included list
;; structures).

(defstruct (layout-binary-tree (:include binary-tree)
                               (:type list))
  x y)



(defun binary-tree-to-layout-binary-tree (tree)
  "
Return a layout-binary-tree homologue to node.
"
  (if (binary-tree-empty-p tree)
      (make-empty-binary-tree)
      (make-layout-binary-tree
       :label (binary-tree-label tree)
       :left  (binary-tree-to-layout-binary-tree (binary-tree-left  tree))
       :right (binary-tree-to-layout-binary-tree (binary-tree-right tree)))))


;; To layout the binary tree, we will do it in two steps.   First we
;; make the layout tree, and setting the y field to the depth of each
;; node.  Then we execute a infix walk of the new tree updating the x
;; field of each node.

(defun layout-node-depth (node depth)
  "
Return a layout-binary-tree homologue to node, with the ordinates of
each node set to their depth.
"
  (if (binary-tree-empty-p node)
      (make-empty-binary-tree)
      (make-layout-binary-tree
       :label (binary-tree-label node)
       :y depth
       :left  (layout-node-depth (binary-tree-left  node) (1+ depth))
       :right (layout-node-depth (binary-tree-right node) (1+ depth)))))



;; Note, incf is a prefix increment, it returns the new-value.
;; Therefore it is easier to start with the predecessor of the first
;; value, and to finally return the last value used.  One could define
;; a postfix increment operator to easily write the code using the
;; other convention.

(defun layout-node-abscissa/inorder (node abscissa)
  "
Sets the abscissa of each node in the subtree NODE to a sequence of
values starting from (1+ ABSCISSA) for the left-most node.
Returns the last abscissa used.
"
  (when (binary-tree-left node)
    (setf abscissa (layout-node-abscissa/inorder (binary-tree-left node) abscissa)))
  (setf (layout-binary-tree-x node) (incf abscissa))
  (when (binary-tree-right node)
    (setf abscissa (layout-node-abscissa/inorder (binary-tree-right node) abscissa)))
  abscissa)


(defun layout-binary-tree-p64 (tree)
  (let ((lobt (layout-node-depth tree 1)))
    (layout-node-abscissa/inorder lobt 0) ; starts from 1; use -1 to start from 0.
    lobt))



(assert (equal (layout-binary-tree-p64  (complete-binary-tree 7))
               '(1
                 (2 (4 NIL NIL 1 3) (5 NIL NIL 3 3) 2 2)
                 (3 (6 NIL NIL 5 3) (7 NIL NIL 7 3) 6 2)
                 4 1)))

(assert (equal (layout-binary-tree-p64   (construct '(n k c a h g e m u p s q) (function string<)))
               '(N (K (C (A NIL NIL 1 4)
                       (H (G (E NIL NIL 3 6)
                             NIL 4 5)
                          NIL 5 4)
                       2 3)
                    (M NIL NIL 7 3)
                    6 2)
                 (U (P NIL
                       (S (Q NIL NIL 10 5)
                          NIL 11 4)
                       9 3)
                    NIL 12 2)
                 8 1)))

(assert (equal (layout-binary-tree-p64   (construct '(n k c a e d g m u p q) (function string<)))
               '(N (K (C (A NIL NIL 1 4)
                       (E (D NIL NIL 3 5)
                          (G NIL NIL 5 5)
                          4 4)
                       2 3)
                    (M NIL NIL 7 3)
                    6 2)
                 (U (P NIL
                       (Q NIL NIL 10 4)
                       9 3)
                    NIL 11 2)
                 8 1)))

;;;; THE END ;;;;
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