1 files changed, 102 insertions, 0 deletions
diff --git a/coding-exercises/2/63.rkt b/coding-exercises/2/63.rkt
new file mode 100644
index 0000000..6a8fd88
--- /dev/null
+++ b/coding-exercises/2/63.rkt
@@ -0,0 +1,102 @@
+#lang racket
+
+(define (entry tree) (car tree))
+(define (left-branch tree) (cadr tree))
+(define (right-branch tree) (caddr tree))
+(define (make-entry entry left right)
+  (list entry left right))
+
+(define (element-of-set? x mset)
+  (cond ((null? mset) false)
+        ((= x (entry mset)) true)
+        ((> x (entry mset))
+         (element-of-set? (right-branch mset)))
+        ((< x (entry mset))
+         (element-of-set? (left-branch mset)))))
+  
+(define (adjoin-set x mset)
+  (cond ((null? mset) (make-tree x '() '()))
+        ((= x (entry mset)) mset)
+        ((< x (entry mset))
+         (make-tree
+           (entry mset)
+           (adjoin-set x (left-branch mset))
+           (right-branch mset)))
+        ((> x (entry))
+         (make-tree
+           (entry mset)
+           (left-branch mset)
+           (adjoin-set x (right-branch mset))))))
+
+(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 '()))
+  
+
+;; depth of a balanced tree with n elements is log n.. right? Because every level you need 2*(level-1) elements, so the total levels is how many time we can do that step until 2*levels > n.
+;; So it is the log a where a is the smallest number that only has factor 2 and > n.
+;; Both procedure 1 and 2 are levelwise reducing the problem.
+;; Only the overhead spent at each level is greater for 1, because append is used in the linear recursive process.
+;; Since at every level we do a linear scan of all left elements, it is n*logn.
+;; 1 is a linear iterative procedure and has almost no overhead at every level except a cons operation so it is logn.
+((lambda ()
+  (define test216a (make-entry 
+                     7
+                     (make-entry 
+                       3
+                       (make-entry
+                         1 '() '())
+                       (make-entry 5 '() '()))
+                      
+                     (make-entry
+                       9
+                       '()
+                       (make-entry
+                         11
+                         '()
+                         '()))))
+  (println (tree->list-1 test216a))
+  (println (tree->list-2 test216a))
+  (newline)
+  (define test216b (make-entry 
+                     3
+                     (make-entry 1 '() '())
+                     (make-entry 
+                       7
+                       (make-entry 5 '() '())
+                       (make-entry
+                         9
+                         '()
+                         (make-entry
+                           11
+                           '()
+                           '())))))
+  (println (tree->list-1 test216b))
+  (println (tree->list-2 test216b))
+  (newline)
+  (define test216c (make-entry 
+                     5
+                     (make-entry 3 (make-entry 1 '() '()) '())
+                     (make-entry 
+                       9
+                       (make-entry 7 '() '())
+                       (make-entry
+                         11
+                         '()
+                         '()))))
+  (println (tree->list-1 test216c))
+  (println (tree->list-2 test216c))
+  (newline)))