position method

@override dynamic position(TreeNode node, double left, double top, double right, double bottom,)

Implementation

@override
position(
    TreeNode node, double left, double top, double right, double bottom) {
  int i0 = 0, i1 = 0;
  num? sumValue, minValue, maxValue, childValue;
  var value = node.value;
  var nodes = node.children!;

  double newRatio, minRatio, alpha, beta;

  while (i0 < nodes.length) {
    final width = right - left;
    final height = bottom - top;

    // Find the next non-empty node.
    do {
      sumValue = nodes[i1++].value;
    } while (sumValue < 0 && i1 < nodes.length);

    minValue = maxValue = sumValue;
    alpha = max(height / width, width / height) / (value * ratio);
    beta = sumValue * sumValue * alpha;
    minRatio = max(maxValue / beta, beta / minValue);

    // Keep adding nodes while the aspect ratio maintains or improves.
    for (; i1 < nodes.length; ++i1) {
      sumValue = sumValue! + (childValue = nodes[i1].value);
      if (childValue < minValue!) minValue = childValue;
      if (childValue > maxValue!) maxValue = childValue;
      beta = sumValue * sumValue * alpha;
      newRatio = max(maxValue / beta, beta / minValue);
      if (newRatio > minRatio) {
        sumValue -= childValue;
        break;
      }
      minRatio = newRatio;
    }

    // Position and record the row orientation.
    var row = TreeNodeBase(children: nodes.sublist(i0, i1), value: sumValue);
    if (width < height)
      dice(row, left, top, right,
          value > 0 ? top += height * sumValue! / value : bottom);
    else
      slice(row, left, top,
          value > 0 ? left += width * sumValue! / value : right, bottom);

    value -= sumValue!;
    i0 = i1;
  }
}