void Split(StaticBinaryTreePartition parent, ushort index)
{
ushort left = parent.GetNewNodeIndex(this.firstChild);
ushort right = parent.GetNewNodeIndex(parent.GetNewChildArrayBaseIndex());
parent.allNodes[index].left = left;
parent.allNodes[index].right = right;
//parent.allNodes may be reallocated in GetNewNodeIndex...
this = parent.allNodes[index];
parent.ReAddChildren(this.firstChild);
}
//add an item to this node
public void Add(IDraw item, float[] itemMin, float[] itemMax, int axis, StaticBinaryTreePartition parent, ushort thisIndex)
{
if (this.min == 0 && this.max == 0)
{
this.min = itemMin[axis];
this.max = itemMax[axis];
}
else
{
this.min = Math.Min(itemMin[axis], this.min);
this.max = Math.Max(itemMax[axis], this.max);
}
if (left == 0)
{
if (childCount != ChildCount)
{
parent.allChildren[(((uint)firstChild) << ChildCountShift) + childCount] = item;
childCount++;
return;
}
Split(parent, thisIndex);
}
float itemCentre = itemMin[axis] + itemMax[axis];
float nodeSplit = this.min + this.max;
ushort goToNode = itemCentre > nodeSplit ? left : right;
//if itemCentre and nodeSplit are equal, then itemCentre > nodeSplit is always false, which
//will mean goToNode == right, so the item will always be added to the right...
//so... If there are more than 'ChildCount' items in the same place, then this can
//cause a stack overflow, because right will keep expanding.
//the solution isn't fast.. but is should keep things ballanced.
if (itemCentre == nodeSplit)
{
//go to the node with fewer children
goToNode =
parent.allNodes[left].CountChildren(parent.allNodes) >
parent.allNodes[right].CountChildren(parent.allNodes) ?
right : left;
}
parent.allNodes[goToNode].Add(item, itemMin, itemMax, axis == 2 ? 0 : axis + 1, parent, goToNode);
}
