Messages in this thread | | | Date | Wed, 25 Aug 2021 21:20:56 +0200 | From | Peter Zijlstra <> | Subject | Re: 答复: 答复 : [PATCH] rbtree: stop iteration early in rb_find_first |
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On Wed, Aug 25, 2021 at 06:26:59PM +0000, Li,Rongqing wrote: > > 10 > / > 5 > \ > 10 > > the above case should not exist. like below, when second 10 is inserted, it should be inserted to right leaf > 10 > / > 5 > > as a result, it should be > > 10 > / \ > 5 10 > > since 10 is not less 10, so new 10 is inserted to right.
You're right that rb_add() does a tail-add for elements it considers equal -- there is actually code in the tree that relies on this.
But that doesn't mean rb_find_first() should go right, it *must*not*, because then it wouldn't find the 'first' aka 'leftmost' instance of the equal elements.
Also, you're only considering building the tree in-order with rb_add(), trees get modified all the time and the pattern Michel gave is perfectly valid (also see rb_prev()).
Sure the snippet is not a balanced tree, but you can construct the pattern as part of a larger tree just fine, just add some elements:
10(b) / \ 5 10(c) \ 10(a)
is a tree that is balanced (remember, RB trees only require the left and right depths to no more than double -- as opposed to AVL trees, which have a tighter constraint). This tree has order: 5, 10(a), 10(b), 10(c). Also note that the tree rotations are stable -- they must be since they do not refence the order function.
As such, if we rb_find_first() for 10, we must find 10(a), the leftmost 10 in the tree.
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