It got rid of overflow but i am doing something wrong in my create function (at least i suspect the problem is there) because i got only few objects displayed compared to what i can see with camera. Please help me further if you find time, thanks.

I can't think of right logic for this:

const std::size_t max_level = 5; // everything above 5 is to slow

void OcTree::create(std::shared_ptr<OcTree> _tree, std::size_t curr_level)
{

std::size_t cnt = 0;
const auto&amp; nodes = pSmgr->getNodes();
for(std::size_t i = 0; i < nodes.size(); ++i)
{
  bool result = _tree->bbox.intersects(nodes[i]->getBBox());
  if(result)
  {
   _tree->vNode.push_back(nodes[i]); // vNodes is vector of objects
   ++cnt;
  }
}
if(cnt <= 1 || curr_level > max_level)
{
  _tree->leaf = true;
  return;
}
else
{
  _tree->vNode.clear();
  for(std::size_t i = 0; i < 8; ++i)
  {
   child[i] = std::shared_ptr<OcTree>(new OcTree(_tree->pSmgr, subdivide(_tree, i)));
   create(child[i], curr_level + 1);
  }
}
}

If you need more info please tell.

It got rid of overflow but i am doing something wrong in my create function (at least i suspect the problem is there) because i got only few objects displayed compared to what i can see with camera. Please help me further if you find time, thanks.

I can't think of right logic for this:

const std::size_t max_level = 5; // everything above 5 is to slow

void OcTree::create(std::shared_ptr<OcTree> _tree, std::size_t curr_level)
{
++levelCnt;
std::size_t cnt = 0;
const auto& nodes = pSmgr->getNodes();
for(std::size_t i = 0; i < nodes.size(); ++i)
{
  bool result = _tree->bbox.intersects(nodes[i]->getBBox());
  if(result)
  {
   _tree->vNode.push_back(nodes[i]); // vNodes is vector of objects
   ++cnt;
  }
}
if(cnt <= 1 || curr_level > max_level)
{
  _tree->leaf = true;
  return;
}
else
{
  _tree->vNode.clear();
  for(std::size_t i = 0; i < 8; ++i)
  {
   child[i] = std::shared_ptr<OcTree>(new OcTree(_tree->pSmgr, subdivide(_tree, i)));
   create(child[i], curr_level + 1);
  }
}
}

If you need more info please tell.