Prolog binary tree node value count

Question

Can anyone help me how to realize predicate in Prolog which counts the all the node numbers of a binary tree?

For example:

tree1(tree(1,
        tree(2,
            tree(3,nil,nil),
            tree(4,nil,nil)),
        tree(5,
            tree(6,nil,nil),
            tree(7,nil,nil))
    )
).

Would return 28. Anyone can help?

Are you sure about 23? Looks like 28 to me, or I am misunderstanding the question. — user1812457
– user1812457, Commented Mar 21, 2013 at 9:39

user1812457 · Accepted Answer · 2013-03-21 09:48:18Z

1

The trivial recursive solution. Not tail-recursive.

treesum(nil, 0).
treesum(tree(X,T1,T2), S) :-
    treesum(T1, S1), treesum(T2, S2),
    S is X+S1+S2.

answered Mar 21, 2013 at 9:48

user1812457

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3 Comments

John Over a year ago

Thank you for the answer, helped me a lot. And how about multiplication? How could I realize that?

John Over a year ago

in this case I can't just change '+' to '*', because I did that and Prolog all the time gave me answer 0(The program GNU Prolog). Because when I traced this multiplication example, in each node it multiplyed also 0. Here is an example of multiplication, just first part, bu all the others are the same: 6 4 Exit: 0 is 3*(0*0) ? `

user1812457 Over a year ago

@John You are joking, right? Here is a riddle: When you add 0, nothing changes. What do you have to multiply to so that nothing changes?

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