在二叉树方案2中查找列表

Question

Write a function in Scheme that receives as input argument list cedulas (identification) of students, a binary search tree with instances of the student structure and returns a list of instances of the student structure, where the identification of the list of students is in the binary search tree.

example:

This is the binary tree

(make-árbol-bin  
(make-estudiante 5 "35889188" "Victor" (make-fecha 10 6 1991) "calle 67 con cra 20" "4444444")  (make-
árbol-bin  
(make-estudiante 2 "6457234" "Manuel" (make-fecha 12 10 1992) "calle 56 con cra 33" "5555555")   (make-árbol-bin 
   (make-estudiante 1 "94252688" "Estela" (make-fecha 20 5 1993) "calle 4 con cra 2" "3333333")    empty    empty)   empty) 
 (make-árbol-bin 
(make-estudiante 7 "34987678" "Juan" (make-fecha 25 3 1995) "calle 34 con cra 12" "6666666")   empty 
 empty) 
)

List identification

(list "94252688" "94888888") 

It should return the following list:

(list (make-estudiante 1 "94252688" "Estela" (make-fecha 20 5 1993) "calle 4 con cra 2" "3333333") ) 

and I did this but I can not make me return the list

(define-struct fecha ( dia mes año))

This is the structure of the student:

(define-struct estudiante ( codigo id nombre fechanaci direccion telefono))

This is the structure of the binary search tree:

(define-struct arbol-bin( estudiante nod-izq nod-der)) 

these are the functions that searches the list in the tree

(define(buscarevd n E)
(cond
[(or(empty? E) (empty? n))false]
[(equal? (first n)(estudiante-id  E)) true]
 [else (buscarevd (rest n) E)]))

(define(buscare n E)
(cond 
 [(empty? E)false]
[(< (car n)(estudiante-id  E)) true]
 [else false]))


(define (member-bt x bt)
 (cond 
 [(empty? bt) false]
[(or(buscarevd x (arbol-bin-estudiante bt))
[else
(member-bt x (arbol-bin-nod-der bt))]))

and this

(member-bt (list "94252688" "94888888") tree)

return

true

As I can do to make me return the list.

Answer 1

Not sure if there is some kind of relationship between nodes which would permit you to skip sub-trees here, so I'd use this kind of approach:

(define (member-bt lst tree)
   (let loop ((tree tree) (res '()))
     (if (null? tree)
         res
         (let* ((node (árbol-bin-estudiante tree))
                (res1 (loop (árbol-bin-nod-der tree) res))
                (res2 (loop (árbol-bin-nod-izq tree) res1)))
           (if (member (estudiante-id node) lst string=?)
               (cons node res2)
               res2)))))

then

(let ((lst '("94888888" "34987678" "94252688")))
  (let ((res (member-bt lst tree)))
    (values res (map estudiante-nombre res))))
=>
'(#<estudiante> #<estudiante>)
'("Estela" "Juan")

Note that this traverses the tree only once, but the order of the students in the result list is not the same as in the initial list. If order needs to be preserved you need to take a different approach, for example:

(define (member-bt-ordered lst tree)
  (define (sub e tree)
    (if (null? tree)
        #f
        (let ((node (árbol-bin-estudiante tree)))
          (if (string=? e (estudiante-id node))
              node
              (or (sub e (árbol-bin-nod-der tree))
                  (sub e (árbol-bin-nod-izq tree)))))))
  (filter values (map (lambda (e) (sub e tree)) lst)))

then

(let ((lst '( "94252688" "94888888" "35889188" "34987678" "6457234")))
  (let ((res (member-bt-ordered lst tree)))
    (values res (map estudiante-nombre res))))
=> 
'(#<estudiante> #<estudiante> #<estudiante> #<estudiante>)
'("Estela" "Victor" "Juan" "Manuel")