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nanoflann: faster check in searchLevel

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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.

Signed-off-by: Luca Bartoli <lucabartoli97@gmail.com>
This commit is contained in:
Luca Bartoli
2025-10-23 18:20:28 +02:00
parent 5714a06a39
commit 2e272106ab
1 changed files with 6 additions and 4 deletions
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10
include/nanoflann.hpp
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@@ -1887,8 +1887,9 @@ class KDTreeSingleIndexAdaptor
DistanceType mindist, distance_vector_t& dists,
const float epsError) const
{
/* If this is a leaf node, then do check and return. */
if ((node->child1 == nullptr) && (node->child2 == nullptr))
// If this is a leaf node, then do check and return.
// If they are equal, both pointers are nullptr.
if (node->child1 == node->child2)
{
DistanceType worst_dist = result_set.worstDist();
for (Offset i = node->node_type.lr.left;
@@ -2314,8 +2315,9 @@ class KDTreeSingleIndexDynamicAdaptor_
DistanceType mindist, distance_vector_t& dists,
const float epsError) const
{
/* If this is a leaf node, then do check and return. */
if ((node->child1 == nullptr) && (node->child2 == nullptr))
// If this is a leaf node, then do check and return.
// If they are equal, both pointers are nullptr.
if (node->child1 == node->child2)
{
DistanceType worst_dist = result_set.worstDist();
for (Offset i = node->node_type.lr.left;