In B-tree insertion algorithm, I see that in order to solve the case in which we need to insert an element to a leaf with 2t-1 elements, we need to do split algorithm to the tree. Something I don't understand is why in the insertion algorithm during the descend in the tree (to the willing point) we split every node with 2t-1 elements, even though I seems useless. for example example
I understand that there is a case in which couple of nodes above the leaf got 2t-1 elements, and in case we want move the median to them we face problem, but why not to give pinpoint solution for that, instead of doing split every time.
correct me if I say something wrong.
We split the full nodes on the way down to the target position because we don't know if we will need to "go back up." You can do it the way you are thinking, where we go down to the target node, split it, and then insert the median of the split into the parent, recursively splitting nodes as needed. But this requires us to go from the root, down to the target, and back up, potentially all the way to the root again. This might be undesirable, e.g. if accessing the nodes twice would be too expensive. In that case, it may be better to go in one pass straight down, where you split any full nodes to anticipate the need for more space.
For a demonstration, you can try inserting 10 into the trees in the middle and on the bottom of your drawing. The tree on the bottom, unsplit, needs to be split all the way to the root in the same way as the middle tree, because the two-pass algorithm didn't leave any space. In the middle tree, inserting 10 still causes a split, but it doesn't extend all the way up because the top two layers of the tree are very spacious.
There is an important caveat, though. Let t be the minimum number of children per node. For the two pass algorithm, the maximum number of children a node can have needs to be at least u = 2t - 1. If it is less, like 2t - 2, then splitting a full node (2t - 3 elements), even with the additional element to insert, will not be able to make two non-deficient nodes. The one pass algorithm requires a higher maximum, u = 2t. This is because the two-pass algorithm always has an element on hand to cancel exactly one deficiency. The one-pass algorithm does not have this ability, as it sometimes splits nodes unnecessarily, so it can't stick the element it's holding into one of the deficiencies. It might not belong there.
