SICP Exercise 2.63: Binary Trees to Ordered Lists

Each of the following two procedures converts a binary tree to a list.

(define (tree->list-1 tree)
  (if (null? tree)
      '()
      (append (tree->list-1 (left-branch tree))
              (cons (entry tree)
                    (tree->list-1 (right-branch tree))))))
(define (tree->list-2 tree)
  (define (copy-to-list tree result-list)
    (if (null? tree)
        result-list
        (copy-to-list (left-branch tree)
                      (cons (entry tree)
                            (copy-to-list (right-branch tree)
                                          result-list)))))
  (copy-to-list tree '()))

1. Do the two procedures produce the same result for every tree? If not, how do the results differ? What lists do the two procedures produce for the trees in figure 2.16?
2. Do the two procedures have the same order of growth in the number of steps required to convert a balanced tree with n elements to a list? If not, which one grows more slowly?

(a) What Do the Procedures Do?
Before we actually look in detail at the internals of the two procedures, let's see how they behave with various trees. It's worth noting that the "Sets as binary trees" section is actually using binary search trees to represent sets, not just binary trees. As a result we'll test the procedures using valid binary search trees:

> (tree->list-1 '())
'()
> (tree->list-2 '())
'()
> (tree->list-1 (make-tree 5 '() '()))
'(5)
> (tree->list-2 (make-tree 5 '() '()))
'(5)
> (tree->list-1 (make-tree 3 (make-tree 2 (make-tree 1 '() '()) '()) '()))
'(1 2 3)
> (tree->list-2 (make-tree 3 (make-tree 2 (make-tree 1 '() '()) '()) '()))
'(1 2 3)
> (tree->list-1 (make-tree 1 '() (make-tree 2 '() (make-tree 3 '() '()))))
'(1 2 3)
> (tree->list-2 (make-tree 1 '() (make-tree 2 '() (make-tree 3 '() '()))))
'(1 2 3)
> (tree->list-1 (make-tree 5
                           (make-tree 3
                                      (make-tree 1
                                                 '()
                                                 (make-tree 2 '() '()))
                                      (make-tree 4 '() '()))
                           (make-tree 6
                                      '()
                                      (make-tree 7 '() '()))))
'(1 2 3 4 5 6 7)
> (tree->list-2 (make-tree 5
                           (make-tree 3
                                      (make-tree 1
                                                 '()
                                                 (make-tree 2 '() '()))
                                      (make-tree 4 '() '()))
                           (make-tree 6
                                      '()
                                      (make-tree 7 '() '()))))
'(1 2 3 4 5 6 7)

As the exercise also asks what the two procedures produce for the trees in figure 2.16, let's try those out too:

> (tree->list-1 (make-tree 7
                           (make-tree 3
                                      (make-tree 1 '() '())
                                      (make-tree 5 '() '()))
                           (make-tree 9
                                      '()
                                      (make-tree 11 '() '()))))
'(1 3 5 7 9 11)
> (tree->list-2 (make-tree 7
                           (make-tree 3
                                      (make-tree 1 '() '())
                                      (make-tree 5 '() '()))
                           (make-tree 9
                                      '()
                                      (make-tree 11 '() '()))))
'(1 3 5 7 9 11)
> (tree->list-1 (make-tree 3
                           (make-tree 1 '() '())
                           (make-tree 7
                                      (make-tree 5 '() '())
                                      (make-tree 9
                                                 '()
                                                 (make-tree 11 '() '())))))
'(1 3 5 7 9 11)
> (tree->list-2 (make-tree 3
                           (make-tree 1 '() '())
                           (make-tree 7
                                      (make-tree 5 '() '())
                                      (make-tree 9
                                                 '()
                                                 (make-tree 11 '() '())))))
'(1 3 5 7 9 11)
> (tree->list-1 (make-tree 5
                           (make-tree 3
                                      (make-tree 1 '() '())
                                      '())
                           (make-tree 9
                                      (make-tree 7 '() '())
                                      (make-tree 11 '() '()))))
'(1 3 5 7 9 11)
> (tree->list-2 (make-tree 5
                           (make-tree 3
                                      (make-tree 1 '() '())
                                      '())
                           (make-tree 9
                                      (make-tree 7 '() '())
                                      (make-tree 11 '() '()))))
'(1 3 5 7 9 11)

Okay, so it looks from the results as if both procedures are performing the equivalent of in-order (or symmetric) tree traversals. Given that these are binary search trees containing numeric values, with the left sub-tree of a node containing only values numerically less than the current node's value and the right-sub-tree of a node containing only values numerically greater than the current node's value, this means that we're expecting both procedures to produce a list containing all of the node values from the tree in numerically ascending order. Let's now look at what each of the two procedures do in turn in order to confirm this.

The first procedure, tree->list-1, processes the child nodes of each node recursively, until it reaches the null children. For any non-null node it appends the tree->list-1 for the left node onto the result of consing the current node's value onto the tree->list-1 for the right node. Assuming that we're dealing with a valid binary search tree then, for any node, this will produce a list containing the current node's value at some index, i such that all the elements in the list before index i are numerically less than the current node's value, while all the elements in the list that are after index i are numerically greater than the current node's value. Given that this is done recursively, we can say that tree->list-1 will indeed produce a list containing all of the node values from the tree in numerically ascending order.

Now let's look at tree->list-1. This procedure (or at least the inner procedure, copy-to-list) is more iterative in nature, building up a result list as it goes along, and stopping once it's run out of tree to process. For any non-null node it recursively calls copy-to-list with the left node as the tree to process, and the result of consing the current node's value onto the list returned by recursively calling copy-to-list with the right node as the tree and the result list as built so far. With the same assumption as before this means that: firstly the head of the list produced by the cons is less than all of the elements in the tail of the list, and secondly all of the elements to be consed onto the head of this list by the call to copy-to-list with the left node as the tree are less than the head of the result list so far. As this is performed recursively, we can also say that tree->list-2 will produce a list containing all of the node values from the tree in numerically ascending order.

How Do the Procedures Grow?
Well, first of all both procedures visit each node once in their traversal of the tree. This means that there's going to be a factor of n in both orders of growth. However, they differ in how they build up the results list...

The first procedure, tree->list-1, uses append to join the sublist produced for the left child node onto the list produced by consing the current node's value onto the sublist produced for the right child node. Now append is an Θ(n) operation, as it must iterate through, in this application, the entirety of the sublist produced for the left child node, and use cons (which we're assuming to be an Θ(1) operation) to append each element in reverse order onto the sublist produced for the right child node. As append is invoked once for each of the n nodes in the tree, tree->list-1 is an Θ(n²) operation.

On the other hand, in each recursive call the second procedure, tree->list-2, uses cons to put the current node's value onto the result list. As noted above we're assuming this to be an Θ(1) operation and, as it's performed once per node, this is performed a total of n times. As a result, tree->list-2 is an Θ(n) operation, and so grows more slowly than tree->list-1.