dredison
How to recurse and build a key to access a child node
Hi,
So I’m new to Elixir and I’m struggling with a recursion question. I have a tree structure (of structs, with ordered lists of children), where each node has a single parent and many children. it can be nested to an arbitrary depth.
tree = %{
id: "a",
children: [
%{
id: "one",
children: [
%{
id: "two"
}
]
},
%{
id: "three",
children: [
%{
id: "four"
},
%{
id: "five"
},
]
},
]
}
I’m trying to write a recursive function that will build an access key to a given node, identified by its ID.
Tree.path_to(tree, id, path \\ [])
``
So for example id "five", would return the following, which I can then use with the kernel's get_in and update_in functions.
[:children, Acess.to(1), :children, Access.to(1)]
I'm just too new to Elixir (and functional programming in general) to work it out so any help would be appreciated!Marked As Solved
aziz
Hi dredison! Was going to give you a partial solution for your learning but I pretty much solved it:
# When parameters appear multiple times they are checked for equality.
def path_to(%{id: id}, id), do: []
def path_to(%{children: [_ | _] = children}, id) do
case subpath_to(children, id) do
{idx, path} -> [:children, idx | path]
_ -> nil
end
end
def path_to(_, _), do: nil
def subpath_to(children, id) do
Enum.reduce_while(children, 0, fn child, idx ->
path = path_to(child, id)
if path, do: {:halt, {idx, path}}, else: {:cont, idx + 1}
end)
end
You still have to do something to get the desired format. 
Tip: When writing recursive functions it always helps to handle the basic cases first and to make sure that there’s a terminating clause. Then you build on that and work your way down to the nested structures and see if you can use the functions that handle the basic cases.
Also Liked
dredison
Hey Aziz,
Amazing, thank you. Works like a charm. It’s also pointed me towards a better understanding of pattern matching and function arguments. Thanks again.
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