```+let t1 = Node ((Node ((Leaf 2), (Leaf 3))),
+               (Node ((Leaf 5),(Node ((Leaf 7),
+                                      (Leaf 11))))))
+
+    .
+ ___|___
+ |     |
+ .     .
+_|__  _|__
+|  |  |  |
+2  3  5  .
+        _|__
+        |  |
+        7  11
+```
+ +Our first task will be to replace each leaf with its double: + +
```+let rec treemap (newleaf:'a -> 'b) (t:'a tree):('b tree) =
+  match t with Leaf x -> Leaf (newleaf x)
+             | Node (l, r) -> Node ((treemap newleaf l),
+                                    (treemap newleaf r));;
+```
+`treemap` takes a function that transforms old leaves into new leaves, +and maps that function over all the leaves in the tree, leaving the +structure of the tree unchanged. For instance: + +
```+let double i = i + i;;
+treemap double t1;;
+- : int tree =
+Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
+
+    .
+ ___|____
+ |      |
+ .      .
+_|__  __|__
+|  |  |   |
+4  6  10  .
+        __|___
+        |    |
+        14   22
+```
+ +We could have built the doubling operation right into the `treemap` +code. However, because what to do to each leaf is a parameter, we can +decide to do something else to the leaves without needing to rewrite +`treemap`. For instance, we can easily square each leaf instead by +supplying the appropriate `int -> int` operation in place of `double`: + +
```+let square x = x * x;;
+treemap square t1;;
+- : int tree =ppp
+Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
+```
+ +Note that what `treemap` does is take some global, contextual +information---what to do to each leaf---and supplies that information +to each subpart of the computation. In other words, `treemap` has the +behavior of a reader monad. Let's make that explicit. + +In general, we're on a journey of making our treemap function more and +more flexible. So the next step---combining the tree transducer with +a reader monad---is to have the treemap function return a (monadized) +tree that is ready to accept any `int->int` function and produce the +updated tree. + +\tree (. (. (f2) (f3))(. (f5) (.(f7)(f11)))) +
```+\f    .
+  ____|____
+  |       |
+  .       .
+__|__   __|__
+|   |   |   |
+f2  f3  f5  .
+          __|___
+          |    |
+          f7  f11
+```
+ +That is, we want to transform the ordinary tree `t1` (of type `int +tree`) into a reader object of type `(int->int)-> int tree`: something +that, when you apply it to an `int->int` function returns an `int +tree` in which each leaf `x` has been replaced with `(f x)`. + +With previous readers, we always knew which kind of environment to +expect: either an assignment function (the original calculator +simulation), a world (the intensionality monad), an integer (the +Jacobson-inspired link monad), etc. In this situation, it will be +enough for now to expect that our reader will expect a function of +type `int->int`. + +
```+type 'a reader = (int->int) -> 'a;;  (* mnemonic: e for environment *)
+```
+ +It's easy to figure out how to turn an `int` into an `int reader`: + +
```+let int2int_reader (x:'a): 'b reader = fun (op:'a -> 'b) -> op x;;
+int2int_reader 2 (fun i -> i + i);;
+- : int = 4
+```
+ +But what do we do when the integers are scattered over the leaves of a +tree? A binary tree is not the kind of thing that we can apply a +function of type `int->int` to. + +
```+let rec treemonadizer (f:'a -> 'b reader) (t:'a tree):('b tree) reader =
+  match t with Leaf x -> reader_bind (f x) (fun x' -> reader_unit (Leaf x'))
+             | Node (l, r) -> reader_bind (treemonadizer f l) (fun x ->
+```
+ +This function says: give me a function `f` that knows how to turn +something of type `'a` into an `'b reader`, and I'll show you how to +turn an `'a tree` into an `'a tree reader`. In more fanciful terms, +the `treemonadizer` function builds plumbing that connects all of the +leaves of a tree into one connected monadic network; it threads the +monad through the leaves. + +
```+# treemonadizer int2int_reader t1 (fun i -> i + i);;
+- : int tree =
+Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
+```
+ +Here, our environment is the doubling function (`fun i -> i + i`). If +we apply the very same `int tree reader` (namely, `treemonadizer +int2int_reader t1`) to a different `int->int` function---say, the +squaring function, `fun i -> i * i`---we get an entirely different +result: + +
```+# treemonadizer int2int_reader t1 (fun i -> i * i);;
+- : int tree =
+Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
+```
+ +Now that we have a tree transducer that accepts a monad as a +parameter, we can see what it would take to swap in a different monad. +For instance, we can use a state monad to count the number of nodes in +the tree. + +
```+type 'a state = int -> 'a * int;;
+let state_unit x i = (x, i+.5);;
+let state_bind u f i = let (a, i') = u i in f a (i'+.5);;
+```
+ +Gratifyingly, we can use the `treemonadizer` function without any +modification whatsoever, except for replacing the (parametric) type +`reader` with `state`: + +
```+let rec treemonadizer (f:'a -> 'b state) (t:'a tree):('b tree) state =
+  match t with Leaf x -> state_bind (f x) (fun x' -> state_unit (Leaf x'))
+             | Node (l, r) -> state_bind (treemonadizer f l) (fun x ->
+                                state_bind (treemonadizer f r) (fun y ->
+                                  state_unit (Node (x, y))));;
+```
+ +Then we can count the number of nodes in the tree: + +
```+# treemonadizer state_unit t1 0;;
+- : int tree * int =
+(Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 13)
+
+    .
+ ___|___
+ |     |
+ .     .
+_|__  _|__
+|  |  |  |
+2  3  5  .
+        _|__
+        |  |
+        7  11
+```
+ +Notice that we've counted each internal node twice---it's a good +exercise to adjust the code to count each node once. + +One more revealing example before getting down to business: replacing +`state` everywhere in `treemonadizer` with `list` gives us + +
```+# treemonadizer (fun x -> [ [x; square x] ]) t1;;
+- : int list tree list =
+[Node
+  (Node (Leaf [2; 4], Leaf [3; 9]),
+   Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))]
+```
+ +Unlike the previous cases, instead of turning a tree into a function +from some input to a result, this transformer replaces each `int` with +a list of `int`'s. + +Now for the main point. What if we wanted to convert a tree to a list +of leaves? + +
```+type ('a, 'r) continuation = ('a -> 'r) -> 'r;;
+let continuation_unit x c = c x;;
+let continuation_bind u f c = u (fun a -> f a c);;
+
+let rec treemonadizer (f:'a -> ('b, 'r) continuation) (t:'a tree):(('b tree), 'r) continuation =
+  match t with Leaf x -> continuation_bind (f x) (fun x' -> continuation_unit (Leaf x'))
+             | Node (l, r) -> continuation_bind (treemonadizer f l) (fun x ->
+                                continuation_bind (treemonadizer f r) (fun y ->
+                                  continuation_unit (Node (x, y))));;
+```
+ +We use the continuation monad described above, and insert the +`continuation` type in the appropriate place in the `treemonadizer` code. +We then compute: + +
```+# treemonadizer (fun a c -> a :: (c a)) t1 (fun t -> []);;
+- : int list = [2; 3; 5; 7; 11]
+```
+ +We have found a way of collapsing a tree into a list of its leaves. + +The continuation monad is amazingly flexible; we can use it to +simulate some of the computations performed above. To see how, first +note that an interestingly uninteresting thing happens if we use the +continuation unit as our first argument to `treemonadizer`, and then +apply the result to the identity function: + +
```+# treemonadizer continuation_unit t1 (fun x -> x);;
+- : int tree =
+Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
+```
+ +That is, nothing happens. But we can begin to substitute more +interesting functions for the first argument of `treemonadizer`: + +
```+(* Simulating the tree reader: distributing a operation over the leaves *)
+# treemonadizer (fun a c -> c (square a)) t1 (fun x -> x);;
+- : int tree =
+Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
+
+(* Simulating the int list tree list *)
+# treemonadizer (fun a c -> c [a; square a]) t1 (fun x -> x);;
+- : int list tree =
+Node
+ (Node (Leaf [2; 4], Leaf [3; 9]),
+  Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))
+
+(* Counting leaves *)
+# treemonadizer (fun a c -> 1 + c a) t1 (fun x -> 0);;
+- : int = 5
+```
+ +We could simulate the tree state example too, but it would require +generalizing the type of the continuation monad to + + type ('a -> 'b -> 'c) continuation = ('a -> 'b) -> 'c;; + +The binary tree monad +--------------------- + +Of course, by now you may have realized that we have discovered a new +monad, the binary tree monad: + +
```+type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;
+let tree_unit (x:'a) = Leaf x;;
+let rec tree_bind (u:'a tree) (f:'a -> 'b tree):'b tree =
+  match u with Leaf x -> f x
+             | Node (l, r) -> Node ((tree_bind l f), (tree_bind r f));;
+```
+ +For once, let's check the Monad laws. The left identity law is easy: + + Left identity: bind (unit a) f = bind (Leaf a) f = fa + +To check the other two laws, we need to make the following +observation: it is easy to prove based on `tree_bind` by a simple +induction on the structure of the first argument that the tree +resulting from `bind u f` is a tree with the same strucure as `u`, +except that each leaf `a` has been replaced with `fa`: + +\tree (. (fa1) (. (. (. (fa2)(fa3)) (fa4)) (fa5))) +
```+                .                         .
+              __|__                     __|__
+              |   |                     |   |
+              a1  .                    fa1  .
+                 _|__                     __|__
+                 |  |                     |   |
+                 .  a5                    .  fa5
+   bind         _|__       f   =        __|__
+                |  |                    |   |
+                .  a4                   .  fa4
+              __|__                   __|___
+              |   |                   |    |
+              a2  a3                 fa2  fa3
+```
+ +Given this equivalence, the right identity law + + Right identity: bind u unit = u + +falls out once we realize that + + bind (Leaf a) unit = unit a = Leaf a + +As for the associative law, + + Associativity: bind (bind u f) g = bind u (\a. bind (fa) g) + +we'll give an example that will show how an inductive proof would +proceed. Let `f a = Node (Leaf a, Leaf a)`. Then + +\tree (. (. (. (. (a1)(a2))))) +\tree (. (. (. (. (a1) (a1)) (. (a1) (a1))) )) +
```+                                           .
+                                       ____|____
+          .               .            |       |
+bind    __|__   f  =    __|_    =      .       .
+        |   |           |   |        __|__   __|__
+        a1  a2         fa1 fa2       |   |   |   |
+                                     a1  a1  a1  a1
+```
+ +Now when we bind this tree to `g`, we get + +
```+           .
+       ____|____
+       |       |
+       .       .
+     __|__   __|__
+     |   |   |   |
+    ga1 ga1 ga1 ga1
+```
+ +At this point, it should be easy to convince yourself that +using the recipe on the right hand side of the associative law will +built the exact same final tree. + +So binary trees are a monad. + +Haskell combines this monad with the Option monad to provide a monad +called a +[SearchTree](http://hackage.haskell.org/packages/archive/tree-monad/0.2.1/doc/html/src/Control-Monad-SearchTree.html#SearchTree) +that is intended to +represent non-deterministic computations as a tree.