将四叉树展平为二维数组

Question

我正在尝试创建一个使用四叉树的游戏。。如图所示，块是树的元素，它们可以有 0 个或 4 个孩子。最大的块在 0 级，它的子块在 1 级，依此类推。对于每个游戏，都有一个 maxLevel，您不能在该 maxLevel 上进一步细分块。让我们称不可分割的块为unit cells。只有叶子有颜色，而非叶子元素的颜色等于 null.

此外，块的子块按此顺序存储在数组中 [upperRight,upperLeft,lowerLeft,lowerRight]。我的目标是将树展平为二维颜色数组（arr[i] 代表行，arr[i][j] 代表棋盘上相应位置的 unit cell 的颜色）。例如，arr[0][0] 处的元素应该具有最大块左上角的 unit cell 的颜色。

这是我到目前为止得到的：

    private Color[][] flatten(Color[][] arr) {
  //base case
   if(this.children.length != 0 && this.level == maxDepth - 1) {
    for (int i = 0; i < arr.length; i++) {
     for (int j = 0; j < arr.length; j++) {

      //I don't know which indices to use. I tried doing some calculations to convert the           coordinates into indices but I coudn't do it.
      arr[?][?] = this.children[j].color;

     }
    }
   }

  //if the current block doesn't have children and it is at maxLevel-1

   else if(this.children.length==0 && this.level == maxDepth - 1){
    for(int i = 0; i < arr.length ; i++){
     for(int j = 0; j < arr.length; j++){

    //same indices problem here
      arr[?][?] = this.color;

     }
    }
   }
  //recursive step
  //I'm doing the recursive call in the order 1,0,2,3 because of the way the children are ordered
   else if(this.children.length != 0 && this.level != maxDepth - 1){
    this.children[1].flatten(arr);
    this.children[0].flatten(arr);
    this.children[2].flatten(arr);
    this.children[3].flatten(arr);
   }

  return arr;
}