CISC 320 Introduction to Algorithms Spring 2014

Transcription

1 CISC 20 Introdution to Algorithms Spring 2014 Leture 9 Red-Blk Trees Courtes of Prof. Lio Li 1

2 Binr Serh Trees (BST) ke[x]: ke stored t x. left[x]: pointer to left hild of x. right[x]: pointer to right hild of x. p[x]: pointer to prent node of x. inr-serh-tree propert: for ever node x: ke[] ke[x] ke[z] where is n node in the left sutree of x, nd z is n node in the right sutree of x. e.g., two vlid BSTs for the kes 2,, 4,,,

3 Inorder-tree-wlk(x) 1. if x nil 2. then inorder-tree-wlk(left[x]);. print ke[x]; 4. inorder-tree-wlk(right[x]); It prints ll elements in monotonill inresing order, in Θ(n) time. BST Serh Serh(T, k) 1. x = root[t]; 2. if x = nil or k = ke[x]. then return x; 4. if k < ke[x]. then return Serh(left[x], k) 6. else return Serh(right[x], k);

4 z () () ()

5 Time: O(h), where h is the tree height. for lned inr tree, h = lg(n) worst-se: h = n

6 Rottion Left rotte x Right rotte x Note: 1. inorder ke ordering is unhnged fter rottion:, x,,, 2. rottion tkes O(1) time. 6

7 Left rotte

8 Left-rotte(T,x) 1. right[x] 2. right[x] left[]. if left[] nil 4. then p[left[]] x. p[] p[x] 6. if p[x] = nil. then root[t] 8. else if x = left[p[x]] 9. then left[p[x]] 10. else right[p[x]] 11. left[] x 12. p[x] x Left rotte x 8

9 Is there mehnism to utomtill rotte whenever the tree is signifintl unlned? 9

10 Red- Blk Trees Red-lk tree is inr serh tree Ever node is either red or lk. Root nd leves (nil) re lk If node is red, then oth hildren re lk All pths from n node x to desendnt lef hve sme numer of lk nodes. Definition: lk-height of node x is the numer of lk nodes (exluding x) on n pth from x to desendnt lef. 10

11 A red-lk tree 12 Intuition: if red-lk tree ontins lk nodes onl, the tree is perfetl lned, i.e., it is omplete inr tree. Presene of red nodes orrupts the lne, ut not muh, euse of the restritions imposed on red nodes. 11

12 Theorem 6. A red-lk tree with n internl nodes hs height t most 2 lg(n+1). proof: ) for n node x, its lk-height denoted s h(x), there re t lest 2 h(x) 1 internl nodes under x (proof indution). ) if the root r hs height h, then h 2 h(r), euse t lest hlf of the nodes on n pth from r to lef must e lk. Aording to ), we hve 2 h(r) 1 n h(r) lg (n+1) h 2 lg(n+1). QED 12

13 Therefore, red-lk tree n never e too off- lned. As result, serhing ke in red-lk tree of n nodes tkes O(2 lg(n+1)) time. How out insert nd delete? More work is needed for these opertions sine the red-lk tree properties need to e mintined. 1

14 Insertion step 1: lote where to insert, vi n unsuessful serh. O(lg n) step 2: insert. O(1) new node is ssigned red olor. wh? - n new node potentill n unlne the tree - if red node gets red hild, RBT is roken. This lerts us to prevent from ontinuousl dding nodes t rnh. - hrd rgument: otherwise the lk-height of the tree is not onserved. step : hek if red-lk-tree properties re dmged. If es, fix it rottions. O(?) 14

15 RB-Insert(T, z) 1. nil[t] 2. x root[t]. while x nil[t] 4. do x. if ke[z] < ke[x] 6. then x left[x]. else x right[x] 8. p[z] 9. If = nil[t] 10. then root[t] z 11. else if ke[z] < ke[] 12. then left[] z 1. else right[] z 14. left[z] nil[t]. right[z] nil[t]. Color[z] RED 1. RB-Insert-Fixup(T, z)

16 Cse 1: red unle E C new z C z A B d D e A B d D e 1. Wh C must e lk? 2. Wh C hs to e red fter B nd D re hnged to lk?. If C is not root, wht shll we do? if C is the root, wht shll we do?

17 Cse 1: red unle C new z C B d D e B d D e A z A 1

18 Cse 2: right hild, lk unle C C d d B A A z z B Does the rottion hnge the lk-height for ffeted pths?? 18

19 Cse : left hild, lk unle C B B d z A C d A z 1. Does this opertion hnge the lkheight for ll ffeted pths? 2. Will it propgte further up? 19

20 Cse 1 Cse Cse 2 z z z z

21 Cse 4, nd 6 re just mirror smmetri to ses 1, 2, nd respetivel, nd n e similrl hndled. 21

22 RB-Insert-Fixup(T, z) 1. while olor[p[z]] = RED 2. do if p[z] = left[p[p[z]]]. then right[p[p[z]]] 4. if olor[] = RED. then olor[p[z]] BLACK 6. olor[] BLACK. olor[p[p[z]]] RED 8. z p[p[z]] 9. else if z = right[p[z]] 10. then z p[z] 11. LEFT-ROTATE(T, z) 12. olor[p[z]] BLACK 1. olor[p[p[z]]] RED 14. RIGHT-ROTATE(T, p[p[z]]). else ( sme s then luse with right nd left exhnged). Color[root[T]] BLACK 22

23 Time nlsis for insertion RB-Insert-Fixup either removes red edge onstnt time (ses 2, nd ) or propgtes red edge one level up (never down), t most to the root, whih is the worst se. As redlk tree of n internl nodes n not e higher thn O(lg n), RB-Insert-Fixup runs in O(lg n) time. Therefore, the totl time is T(n) = O(lg n) + O(lg n) = O(lg n) 2