Homework 4 Due Thursday Oct 7. CLRS 12-4 (number of binary trees) CLRS (rb insert implementation)

Transcription

1 Homework 4 Due Thurday Oct 7 CLRS 12-4 (number of binary tree) CLRS (rb inert imlementation) 1

2 Chater 13: Red-Black Tree A red-black tree i a node-colored BST. Each node i colored either black or red. The following ecial rule aly: 1. The root i alway black. 2. A nil i conidered to be black. Thi mean that every non-nil node ha two children. 3. Black Children Rule: The children of each red node are black. 4. Black Height Rule: For each node v, there eit an integer bh(v) uch that each downward ath from v to a nil ha eactly bh(v) black real (i.e. non-nil) node. Call thi quantity the black height of v. We define the black height of an RB tree to be the black height of it root. 2

3 An Eamle NIL 2 NIL NIL : black node : red node NIL NIL NIL NIL NIL 15 NIL 3

4 RB Tree Are Balanced Lemma A Let T be an RB tree having ome n 1 node. Then the height of T i at mot 2 lg(n + 1) 1. Proof Let h be the height of T. Let v 0, v 1,..., v h+1 be an arbitrary length h downward ath from the root to a nil, where v 0 i the root, v h i the leaf, and v h+1 i the nil. v 0 i black and v h+1 i black. The number of red node among v 1,..., v h i maimized when for all odd i v i i red. So, the number of red node i at mot h/2. Thi mean that the number of black one among v 1,..., v h i at leat h h/2. Thu, bh(t ) h/2 4

5 Proof (cont d) If a node ha a real black child, then it ha another child. Thi mean that the tree contain a comlete binary tree of height bh(t ) 1, coniting olely of real black node. The number of node in the comlete binary tree i 2 bh(t ) 1 black node in T. Thi number i at mot n. So, we have lg(n + 1) bh(t ) h/2. By olving thi we have h 2 lg(n + 1). 5

6 Oeration on RB Tree We will tudy two oeration, inertion and deletion. The two oeration make ue of two oeration, Left-Rotate and Right-Rotate. 6

7 Left-Rotate and Right-Rotate y right-rotate at y γ α y α β left-rotate at β γ Rotation do not break the BST-roerty. 7

8 1. Inertion Suoe that we want to inert a node into an RB tree T. To do thi, we inert a a red node uing the inertion algorithm for BST and then reolve violation of the coloring rule. The ecetion i when the tree T i emty. Then we color black. Will thi oeration violate any rule? 8

9 Will thi oeration violate any rule? Becaue the node i red, the Black Children Rule may be violated. The violation haen when the arent of i red. The other rule are not violated. 9

10 Enforcing the Black Children Rule After Inertion Let be the arent of. Aume i red; otherwie there i no violation. Since i red, it cannot be the root. So, let g be the grand arent of. Let u be the ibling of. Let be the ibling of, which can be nil. Since there wa no violation before inertion of, g i black and i black. However, u can be either red or black. We aume that i the left child of g. The treatment in the cae when i the right child i imilar. We conider two cae, u i black and u i red. 10

11 (Cae 1) color[u] = black Firt, if i not the left child of, then left-rotate at and wa the role of and that of. Thi reerve bh at g oition. Net, right-rotate at g, then wa the color of and that of g. Alo, at the end the node at g oition i black. So, there i no violation any more. 11

12 g u new left rotate at new e g u new d e new d d new e g u g right rotate at g d e u d e g wa color of & g d e u violation reolved 12

13 (Cae 2) color[u] = red In thi cae, we color both and u black and color g red. Thi eliminate the violation of the Black Children Rule between and, but may introduce violation, which i between g and it arent. So, we may be back to quare one, but the location of the violation, if introduced, i two level cloer to the root. Thu, the bad ituation doe not reeat more than the height of the tree. 13

14 g u g u violation may occur 14

15 What will haen at the end? Either Cae 1 hold or the reolution for Cae 2 doe not introduce violation. What other condition might be violated? 15

16 It i when the reolution for Cae 2 eliminate violation of the Black Children Rule but turn the root red. 16

17 Eamle 1: g u 8 g 11 8 g cae 2 alie u cae 1 alie u wa color

18 Eamle 2: 10 1 wa color left rotate right rotate 20 left rotate wa color

19 2. Deletion of a node z We firt aly BST-Delete. In the cae when a node i coied to z (the ucceor of z come to z oition), color the new one by the color of z. The deletion routine give back the ointer,, to a node where the actual elimination took lace. There are two oibilitie: (1) There wa a leaf at oition and i a nil (2) The node who wa at oition had a unique child and now thi unique child i at oition. In the latter cae, the unique child i red. Why? 19

20 In the latter cae, the unique child i red. Aume otherwie. Why? Then the black height of the other ubtree, which i a nil, i one, while the black height of the unique ubtree i at leat two. That mean that the Black Height Rule i already violated. 20

21 So, the latter cae will not create violation. Thu, we will conider only the cae when there wa a leaf at i oition. Furthermore, if the leaf that ha been eliminated i red, then the elimination doe not introduce violation. So, we aume that the leaf i black. 21

22 Reolving the Black Height Rule Violation Let w be a node. We ay that Few(w) hold if the following condition are atified: The Black Height Rule i violated in the red-black tree. T w, the ubtree rooted at w, i a red-black tree without coloring rule violation, ecet the rule about the color of the root. If bh(w) were bh(w) + 1 then the Black-Height Rule would be atified for all the node outide T w 22

23 Some Proertie of Few( ) The condition Few initially hold at the nil who relacing. If Few(w) hold and w i red, then coloring w black reolve violation. So, we will conider the cae when w i black. If Few(w) hold then w cannot be the root. 23

24 Aume that i the left child of it arent. The cae in which i the right child can be olved imilarly. Let be ibling. Let A and B be the left child and the right child of, reectively. Thee two node eit ince we ve eliminated a real black node. Here we can reroce the tree o that the following two condition are atified: 1. the ibling of i black and 2. if the left child of the ibling of i red, then the right child of the ibling of i red. 24

25 -1 A B

26 Etablihing Proerty #1 We need to make ure that the ibling of i black. Suoe that i red. Then it arent and it children are all black. That i,, A, and B are black. Suoe we left-rotate at and then wa the color of and. Then, the ibling of i now A and i black; the bh-value i unchanged for, A, and B, o Few() till hold; the deth of, i.e. the location of Few, i increaed by 1; and the arent of i red. 26

27 1 A B 0 0 left rotate wa color 1 A 0 B 0 27

28 Etablihing Proerty #2 We need to enure that if the left child of the ibling of i red, then the right child of the ibling of i red. Suoe that A i red and B i black. Let C and D be the children of A. Then thee two are black. Suoe we right-rotate at and wa the color of A and. Then, A become the ibling of ; C and become the left child and the right child of A, reectively; the bh-value i reerved for, C, D, B, and the ibling of, o Few() till hold; the left child of ibling i now black. 28

29 right rotate wa color 1 A C D 0 0 B 0 1 C 1 A D B C 0 A D B

30 Reolution After Enforcing the Condition In the cae when A and B are both black, we do the following. If i black, then we color red. Then, the downward ath going through loe one black node. Thu, the location of the Few condition move one level u to. If i red, we color black and color red. Then the Few condition diaear. 30

31 A B A B A B A B 31

32 The Remaining Cae A B tatu B B B done R B B done R R B rohibited B R B rohibited R R R to be done R B R to be done B R R to be done B B R to be done We left-rotate at. Furthermore, in the cae of (R,R,R) we color red and color and B black and in the cae of (B,R,R) and (B,B,R) we color B black. 32

33 B B A A A B B B A A B B A A A B B B A A B A 33