treetraversal.cpp

#include <vector>
#include <string>
#include <iostream>

enum class _tree_return {
    Continue = 0,
    Prune,
    Break
};

template<typename T, typename F1, typename F2, typename F3>
_tree_return _for_tree(T initial, F1 condition, F2 branch, F3 visit) {
    _tree_return result = visit(initial);
    if(result == _tree_return::Break) return _tree_return::Break;

    if(result != _tree_return::Prune) {
        for(T subnode : branch(initial)) {
            if(condition(subnode)) {
                _tree_return result = _for_tree(subnode, condition, branch, visit);
                if(result == _tree_return::Break) return _tree_return::Break;
            }
        }
    }
    return _tree_return::Continue;
}

#define tree_break return _tree_return::Break
#define tree_prune return _tree_return::Prune
#define tree_continue return _tree_return::Continue

                                                          //v-- semicolon to not allow you to get the return value here
#define for_tree(XName, Xinitial, Condition, Branch, Visit) ;_for_tree(Xinitial, \
[&](decltype(Xinitial) XName){ return Condition; }, \
[&](decltype(Xinitial) XName){ return std::vector<decltype(Xinitial)>Branch; }, \
[&](decltype(Xinitial) XName){ Visit; return _tree_return::Continue; })
//excuse the use of a std::vector in there, I guess you cant return an initialize_list from a lambda
//that wouldn't really be an issue if this was implemented at the language level instead of hacked together from lambdas and macros

struct Node {
    Node* left = NULL;
    Node* right = NULL;
    std::string value;
    Node(std::string value):value(value){}
};

int main() {
    //syntax is a little uglier than it could be if it was native
    
    //imperative tree sample
    for_tree(x, std::string(""), x.size()<=8, ({x+"a", x+"b", x+"c"}), {
        std::cout << x << std::endl;
    });
    
    //tree structure sample
    Node mytree("root");
    mytree.left = new Node("left");
    mytree.right = new Node("right");
    mytree.left->left = new Node("leftleft");
    
    for_tree(x, &mytree, x != NULL, ({x->left, x->right}), {
        std::cout << x->value << std::endl;
    });

    return 0;
}