Question

I am trying to implement an inorder traversal which in every stage I’ll get the current node.
For example:

?- getnodesinorder(tree(1,nil,nil),X).
X = 1 ;
false.

?- getnodesinorder(tree(5,tree(4,tree(1,nil,tree(3,tree(2,nil,nil),nil)),nil),tree(6,nil,nil)),X).
X = 1 ;
X = 2 ;
X = 3 ;
X = 4 ;
X = 5 ;
X = 6 ;
false. 

I’ve tried the next code:

getnodesinorder(tree(CurrentNode,nil,nil), CurrentNode).
getnodesinorder(tree(X, Left, nil), CurrentNode) :-
    getnodesinorder(Left, _ ),
    CurrentNode is X.
getnodesinorder(tree(X, nil, Right), CurrentNode) :-
    CurrentNode is X,
    getnodesinorder(Right, _ ).
getnodesinorder(tree(X, Left, Right), CurrentNode) :-
    getnodesinorder(Left, _ ),
    CurrentNode is X,
    getnodesinorder(Right, _ ).

So of course the base (1st example works) but when trying to run the 2nd one I get

X=5;
false

as result. Why is that?

Answer 1

The error occurs because you’re processing the left and right subtrees, but not doing anything with the values in them: getnodesinorder(Left, _ ) just throws them away. So, your predicate only returns the top element.

Here’s how you do an in-order traversal:

inorder(tree(_,L,_), X) :- inorder(L,X).
inorder(tree(X,_,_), X).
inorder(tree(_,_,R), X) :- inorder(R,X).

Example query:

?- inorder(tree(5,tree(4,tree(1,nil,tree(3,tree(2,nil,nil),nil)),nil),tree(6,nil,nil)),X).
X = 1 ;
X = 2 ;
X = 3 ;
X = 4 ;
X = 5 ;
X = 6 ;
false.