nanoflann: faster check in searchLevel

[lucabart97]

Instead of separately confirming if both children are nullptr, compare the child pointers' values directly. This technique is valid because, in a properly structured search tree, the two child pointers will only be equal if both are nullptr.

This change reduces the operation from three comparisons to just one.

[jlblancoc]

@lucabart97 wow, that's smart.
Probably just ripping a nanosecond per iteration... Or maybe not even that because at the microarchitecture level all CPUs have a special op to check for a register to be zero, so it should be cheaper than a A==B comparison.

Anyways, I'm triggering the CI tests, if they all pass, I'm glad to merge. Before, please add a small comment to those lines to make clear to future readers that it's actually checking for nullptr.

Thanks! 🙏

[lucabart97]

Ok, done✈️

[qq422216549]

case 1:child1==nullptr child2==nullptr
case 2:child1==nullptr child2 != nullptr
case 3:child1 != nullptr child2==nullptr
case 4:child1 != nullptr child2 != nullptr
Actually case2 and case3 Will not appear
so just compare child1 is Enough

if(node->child1==nullptr)
{
......
}