Ocaml - tree preorder/postorder/inorder
0
I have that definition of trees given in OCaml
type 'a tree = Node of 'a * 'a tree list;;
let rec fold_tree f (Node (x,l)) =
f x (map (fold_tree f) l);;
Can somebody help me how can I write for example preorder in use of fold_tree (without additional recursion). I know how I can do this without fold_tree but this makes me problem
So far I have got this:
let preorder t =
fold_tree (fun x l ->
(fold_left(fun acc h -> h@acc) x l ) ) t;;
but ocaml consider t as tree list...
tree ocaml preorder
asked Nov 12 '18 at 20:40
mvxxxmvxxx
556
|
show 2 more comments
0
I have that definition of trees given in OCaml
type 'a tree = Node of 'a * 'a tree list;;
let rec fold_tree f (Node (x,l)) =
f x (map (fold_tree f) l);;
Can somebody help me how can I write for example preorder in use of fold_tree (without additional recursion). I know how I can do this without fold_tree but this makes me problem
So far I have got this:
let preorder t =
fold_tree (fun x l ->
(fold_left(fun acc h -> h@acc) x l ) ) t;;
but ocaml consider t as tree list...
tree ocaml preorder
asked Nov 12 '18 at 20:40
mvxxxmvxxx
556
2
What have you written so far, and why does it not seem to be working?
– Jeffrey Scofield
Nov 12 '18 at 20:56
@JeffreyScofield yes, you have right, I mentioned problem but I did not say anything about my trying. I edited a post.
– mvxxx
Nov 12 '18 at 21:31
1
Did you not take anything from the answer to your last question? If you follow those advices, what happens then?
– glennsl
Nov 12 '18 at 21:33
ignoring open and hd does not matter there. But if it comes to another advices, or I didnt understand them or there does not matter too in that situation
– mvxxx
Nov 12 '18 at 21:45
You don't think type annotations matter for understanding the cause of a type error?
– glennsl
Nov 12 '18 at 21:56
|
show 2 more comments
0
0
0
I have that definition of trees given in OCaml
type 'a tree = Node of 'a * 'a tree list;;
let rec fold_tree f (Node (x,l)) =
f x (map (fold_tree f) l);;
Can somebody help me how can I write for example preorder in use of fold_tree (without additional recursion). I know how I can do this without fold_tree but this makes me problem
So far I have got this:
let preorder t =
fold_tree (fun x l ->
(fold_left(fun acc h -> h@acc) x l ) ) t;;
but ocaml consider t as tree list...
tree ocaml preorder
asked Nov 12 '18 at 20:40
mvxxxmvxxx
556
I have that definition of trees given in OCaml
type 'a tree = Node of 'a * 'a tree list;;
let rec fold_tree f (Node (x,l)) =
f x (map (fold_tree f) l);;
Can somebody help me how can I write for example preorder in use of fold_tree (without additional recursion). I know how I can do this without fold_tree but this makes me problem
So far I have got this:
let preorder t =
fold_tree (fun x l ->
(fold_left(fun acc h -> h@acc) x l ) ) t;;
but ocaml consider t as tree list...
tree ocaml preorder
tree ocaml preorder
asked Nov 12 '18 at 20:40
mvxxxmvxxx
556
asked Nov 12 '18 at 20:40
mvxxxmvxxx
556
edited Nov 12 '18 at 21:30
mvxxx
asked Nov 12 '18 at 20:40
mvxxxmvxxx
556
asked Nov 12 '18 at 20:40
mvxxxmvxxx
556
asked Nov 12 '18 at 20:40
mvxxxmvxxx
556
556
2
What have you written so far, and why does it not seem to be working?
– Jeffrey Scofield
Nov 12 '18 at 20:56
@JeffreyScofield yes, you have right, I mentioned problem but I did not say anything about my trying. I edited a post.
– mvxxx
Nov 12 '18 at 21:31
1
Did you not take anything from the answer to your last question? If you follow those advices, what happens then?
– glennsl
Nov 12 '18 at 21:33
ignoring open and hd does not matter there. But if it comes to another advices, or I didnt understand them or there does not matter too in that situation
– mvxxx
Nov 12 '18 at 21:45
You don't think type annotations matter for understanding the cause of a type error?
– glennsl
Nov 12 '18 at 21:56
|
show 2 more comments
2
What have you written so far, and why does it not seem to be working?
– Jeffrey Scofield
Nov 12 '18 at 20:56
@JeffreyScofield yes, you have right, I mentioned problem but I did not say anything about my trying. I edited a post.
– mvxxx
Nov 12 '18 at 21:31
1
Did you not take anything from the answer to your last question? If you follow those advices, what happens then?
– glennsl
Nov 12 '18 at 21:33
ignoring open and hd does not matter there. But if it comes to another advices, or I didnt understand them or there does not matter too in that situation
– mvxxx
Nov 12 '18 at 21:45
You don't think type annotations matter for understanding the cause of a type error?
– glennsl
Nov 12 '18 at 21:56
2
2
What have you written so far, and why does it not seem to be working?
– Jeffrey Scofield
Nov 12 '18 at 20:56
What have you written so far, and why does it not seem to be working?
– Jeffrey Scofield
Nov 12 '18 at 20:56
@JeffreyScofield yes, you have right, I mentioned problem but I did not say anything about my trying. I edited a post.
– mvxxx
Nov 12 '18 at 21:31
@JeffreyScofield yes, you have right, I mentioned problem but I did not say anything about my trying. I edited a post.
– mvxxx
Nov 12 '18 at 21:31
1
1
Did you not take anything from the answer to your last question? If you follow those advices, what happens then?
– glennsl
Nov 12 '18 at 21:33
Did you not take anything from the answer to your last question? If you follow those advices, what happens then?
– glennsl
Nov 12 '18 at 21:33
ignoring open and hd does not matter there. But if it comes to another advices, or I didnt understand them or there does not matter too in that situation
– mvxxx
Nov 12 '18 at 21:45
ignoring open and hd does not matter there. But if it comes to another advices, or I didnt understand them or there does not matter too in that situation
– mvxxx
Nov 12 '18 at 21:45
You don't think type annotations matter for understanding the cause of a type error?
– glennsl
Nov 12 '18 at 21:56
You don't think type annotations matter for understanding the cause of a type error?
– glennsl
Nov 12 '18 at 21:56
|
show 2 more comments
1 Answer
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OCaml version 4.02.3
# type 'a tree = Node of 'a * 'a tree list;;
type 'a tree = Node of 'a * 'a tree list
# let rec fold_tree f (Node (x,l)) =
f x (List.map (fold_tree f) l);;
val fold_tree : ('a -> 'b list -> 'b) -> 'a tree -> 'b = <fun>
# let preorder t =
fold_tree (fun x l ->
(List.fold_left(fun acc h -> h@acc) x l ) ) t;;
val preorder : 'a list tree -> 'a list = <fun>
You are using x as start value for the List.fold_left and append values to it. That makes x a 'a list and therefore t must be 'a list tree. Change x to [x] and you get:
# let preorder t =
fold_tree (fun x l ->
(List.fold_left(fun acc h -> h@acc) [x] l ) ) t;;
val preorder : 'a tree -> 'a list = <fun>
answered Nov 19 '18 at 13:58
Goswin von BrederlowGoswin von Brederlow
2,817724
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OCaml version 4.02.3
# type 'a tree = Node of 'a * 'a tree list;;
type 'a tree = Node of 'a * 'a tree list
# let rec fold_tree f (Node (x,l)) =
f x (List.map (fold_tree f) l);;
val fold_tree : ('a -> 'b list -> 'b) -> 'a tree -> 'b = <fun>
# let preorder t =
fold_tree (fun x l ->
(List.fold_left(fun acc h -> h@acc) x l ) ) t;;
val preorder : 'a list tree -> 'a list = <fun>
You are using x as start value for the List.fold_left and append values to it. That makes x a 'a list and therefore t must be 'a list tree. Change x to [x] and you get:
# let preorder t =
fold_tree (fun x l ->
(List.fold_left(fun acc h -> h@acc) [x] l ) ) t;;
val preorder : 'a tree -> 'a list = <fun>
answered Nov 19 '18 at 13:58
Goswin von BrederlowGoswin von Brederlow
2,817724
add a comment |
0
OCaml version 4.02.3
# type 'a tree = Node of 'a * 'a tree list;;
type 'a tree = Node of 'a * 'a tree list
# let rec fold_tree f (Node (x,l)) =
f x (List.map (fold_tree f) l);;
val fold_tree : ('a -> 'b list -> 'b) -> 'a tree -> 'b = <fun>
# let preorder t =
fold_tree (fun x l ->
(List.fold_left(fun acc h -> h@acc) x l ) ) t;;
val preorder : 'a list tree -> 'a list = <fun>
You are using x as start value for the List.fold_left and append values to it. That makes x a 'a list and therefore t must be 'a list tree. Change x to [x] and you get:
# let preorder t =
fold_tree (fun x l ->
(List.fold_left(fun acc h -> h@acc) [x] l ) ) t;;
val preorder : 'a tree -> 'a list = <fun>
answered Nov 19 '18 at 13:58
Goswin von BrederlowGoswin von Brederlow
2,817724
add a comment |
0
0
0
OCaml version 4.02.3
# type 'a tree = Node of 'a * 'a tree list;;
type 'a tree = Node of 'a * 'a tree list
# let rec fold_tree f (Node (x,l)) =
f x (List.map (fold_tree f) l);;
val fold_tree : ('a -> 'b list -> 'b) -> 'a tree -> 'b = <fun>
# let preorder t =
fold_tree (fun x l ->
(List.fold_left(fun acc h -> h@acc) x l ) ) t;;
val preorder : 'a list tree -> 'a list = <fun>
You are using x as start value for the List.fold_left and append values to it. That makes x a 'a list and therefore t must be 'a list tree. Change x to [x] and you get:
# let preorder t =
fold_tree (fun x l ->
(List.fold_left(fun acc h -> h@acc) [x] l ) ) t;;
val preorder : 'a tree -> 'a list = <fun>
answered Nov 19 '18 at 13:58
Goswin von BrederlowGoswin von Brederlow
2,817724
OCaml version 4.02.3
# type 'a tree = Node of 'a * 'a tree list;;
type 'a tree = Node of 'a * 'a tree list
# let rec fold_tree f (Node (x,l)) =
f x (List.map (fold_tree f) l);;
val fold_tree : ('a -> 'b list -> 'b) -> 'a tree -> 'b = <fun>
# let preorder t =
fold_tree (fun x l ->
(List.fold_left(fun acc h -> h@acc) x l ) ) t;;
val preorder : 'a list tree -> 'a list = <fun>
You are using x as start value for the List.fold_left and append values to it. That makes x a 'a list and therefore t must be 'a list tree. Change x to [x] and you get:
# let preorder t =
fold_tree (fun x l ->
(List.fold_left(fun acc h -> h@acc) [x] l ) ) t;;
val preorder : 'a tree -> 'a list = <fun>
answered Nov 19 '18 at 13:58
Goswin von BrederlowGoswin von Brederlow
2,817724
answered Nov 19 '18 at 13:58
Goswin von BrederlowGoswin von Brederlow
2,817724
answered Nov 19 '18 at 13:58
Goswin von BrederlowGoswin von Brederlow
2,817724
answered Nov 19 '18 at 13:58
Goswin von BrederlowGoswin von Brederlow
2,817724
2,817724
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add a comment |
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2
What have you written so far, and why does it not seem to be working?
– Jeffrey Scofield
Nov 12 '18 at 20:56
@JeffreyScofield yes, you have right, I mentioned problem but I did not say anything about my trying. I edited a post.
– mvxxx
Nov 12 '18 at 21:31
1
Did you not take anything from the answer to your last question? If you follow those advices, what happens then?
– glennsl
Nov 12 '18 at 21:33
ignoring open and hd does not matter there. But if it comes to another advices, or I didnt understand them or there does not matter too in that situation
– mvxxx
Nov 12 '18 at 21:45
You don't think type annotations matter for understanding the cause of a type error?
– glennsl
Nov 12 '18 at 21:56