what to do when node was already in close in A* algorithm
Get the square on the open list which has the lowest score. Let’s call this square S.
Remove S from the open list and add S to the closed list.
For each square T in S’s walkable adjacent tiles:
If T is in the closed list: Ignore it.
If T is not in the open list: Add it and compute its score.
If T is already in the open list: Check if the F score is lower when we use the current generated path to get there. If it is, update its score and update its parent as well.
