<p>In <a href="https://github.com/mlpack/mlpack/pull/747#discussion_r74272371">src/mlpack/core/tree/spill_tree/spill_tree_impl.hpp</a>:</p>
<pre style='color:#555'>> + }
> +
> + std::vector<size_t> leftPoints, rightPoints;
> + // Split the node.
> + overlappingNode = SplitPoints(tau, rho, points, leftPoints, rightPoints);
> +
> + // We don't need the information in points, so lets clean it.
> + std::vector<size_t>().swap(points);
> +
> + // Now we will recursively split the children by calling their constructors
> + // (which perform this splitting process).
> + left = new SpillTree(this, leftPoints, tau, maxLeafSize, rho);
> + right = new SpillTree(this, rightPoints, tau, maxLeafSize, rho);
> +
> + // Update count number, to represent the number of descendant points.
> + count = left->NumDescendants() + right->NumDescendants();
</pre>
<p>Hi <a href="https://github.com/rcurtin" class="user-mention">@rcurtin</a> , thanks for your comments.<br>
Yes, I have thought of that approach that you mention, and many similar options. I tried to explain the problem in the previous post, but it is a bit difficult to formulate.<br>
Suppose I implement your approach. And we have a tree of 2 levels of overlapping nodes:</p>
<pre><code> N
|''''''''''\
N_1 N_2
| \ | \
N_11 N_12 N_21 N_22
</code></pre>
<p>The problem is how to organize the points in the second level, in the nodes <code>N_21</code> and nodes <code>N_22</code>.<br>
Suppose we want to get the <code>i-th</code> descendant point of the node N, included in N_2, but not included in N_1.<br>
This means when <code>i >= N_1.NumDescendants()</code>.</p>
<p>Let U: points held in both N_1 and N_2.</p>
<p>As, N_2 is splitted in N_21 and N_22 by another different hyperplane, some points of U will be included in N_21 and some points of U will be included in N_22.</p>
<p>If <code>i < N_1.NumDescendants() + N_21.NumDescendants()</code>, we will call: <code>N_21.Descendant( i - N_1.NumDescendants())</code></p>
<p>As N_21 includes some points from U, we can repeat a point that was included in the node N_1. </p>
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