stevensonmt

stevensonmt

I simply could not figure out a way to solve this problem without using a GenServer to keep the state of the running sum. My solution with GenServer is below, but I’m wondering if anyone could suggest a way to do it without a GenServer, which felt like overkill.

  use GenServer 
  
  def init(tally), do: {:ok, tally}

  def handle_cast({:add, n}, tally), do: {:noreply, n + tally}

  def handle_call(:tally, _from, tally), do: {:reply, tally, tally}


  @spec convert_bst(root :: TreeNode.t | nil) :: TreeNode.t | nil
  def convert_bst(root) do
    {:ok, pid} = GenServer.start(__MODULE__, 0)
    do_convert(root, pid)
  end
 
  def do_convert(nil, tally), do: nil
  def do_convert(node, tally) do
    right = do_convert(node.right, tally)
    GenServer.cast(tally, {:add, node.val})
    new_val = GenServer.call(tally, :tally)
    left = do_convert(node.left, tally)
    %TreeNode{ val: new_val, left: left, right: right}
  end

Marked As Solved

al2o3cr

al2o3cr

The GenServer lets do_convert do two things:

  • return a new tree node
  • as a side-effect, update the running total

To remove it, you’ll need to transform that side-effect into a part of the return value.

Consider Map.pop for inspiration; it returns a {popped_value, updated_map} tuple. You can combine it with rebinding to get code that looks almost like mutability:

things = %{a: 1, b: 2, c: 3}

{c_value, things} = Map.pop(things, :c)

A similar transformation on do_convert would make it return a tuple of {tree_node, updated_tally}.

Also Liked

stevensonmt

stevensonmt OP

Many thanks. I knew there must be some way to keep that “state” without using a GenServer. My first issue had actually been figuring out how to use the left node value as the tally for the grandparent node calculation. I only figured that out after changing to GenServer which made the steps in the process clearer to me. Changing to returning tuples was trivial. Thanks again.

  def convert_bst(root) do
    do_convert(root, 0) |> elem(0)
  end
 
  def do_convert(nil, tally), do: {nil, tally}
  def do_convert(node, tally) do
    {right, interim_tally} = do_convert(node.right, tally)
    new_val = interim_tally + node.val
    {left, final_tally} = do_convert(node.left, new_val)
    {%TreeNode{ val: new_val, left: left, right: right}, final_tally}
  end
Aetherus

Aetherus

I came up with a similar solution, and it passed the tests:

defmodule Solution do
  @spec convert_bst(root :: TreeNode.t | nil) :: TreeNode.t | nil  
  def convert_bst(root) do
    {converted, _} = do_convert_bst(root, 0)
    converted
  end
    
  defp do_convert_bst(nil, _carry), do: {nil, 0}
  defp do_convert_bst(node, carry) do
    {right, sum_r} = do_convert_bst(node.right, carry)
    carry = carry + node.val + sum_r
    {left, sum_l} = do_convert_bst(node.left, carry)
    {%TreeNode{left: left, right: right, val: carry}, node.val + sum_l + sum_r}
  end
end

carry is the sum of the values of all the nodes that have already been visited that have values greater than the next node to visit.

joaoevangelista

joaoevangelista

I have no ideia how to implement such algorithm, but!
Genservers are an abstraction over a looping receiver, so you can try to implement it with recursion, keeping the state as a parameter and keep passing it. Start with the base cases that returns a value instead of recusing. I bet it will look closely to your current implementation,but without the genserver specifics

Last Post!

stackcats

stackcats

There are many problems in Leetcode with the topic Design that you can’t do without using GenServer :joy:

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