storage/spanlatch: catch identical upper bound on recomputation durin…

…g removal

This change modifies `adjustUpperBoundOnRemoval` to avoid a degenerate
case in element removal where all intervals have the same end key. In
this case, we would previously adjust the upper bound of every node from
the root of the tree to the node that the interval was being removed
from. We now check whether removing the element with the largest end key
is actually changing the upper bound of the node. If there are other
elements with the same end key then this is not the case and we can
avoid repeat calls to `adjustUpperBoundOnRemoval` while traversing back
up the tree.

This came up while profiling a benchmark that was giving suprising
results.

Release note: None

pkg/storage/spanlatch/interval_btree.go

@@ -626,8 +626,9 @@ func (n *node) adjustUpperBoundOnRemoval(la *latch, child *node) bool {
 		}
 	}
 	if n.max.compare(up) == 0 {
+		// up was previous upper bound of n.
 		n.max = n.findUpperBound()
-		return true
+		return n.max.compare(up) != 0
 	}
 	return false
 }