13 Red-Black Trees

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13. Red-Black Tree

Hsu, Lih-Hsing


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Chapter 13 P.2

Computer Theory Lab.

One of many search-tree schemes thatare balanced in order to guarantee

that basic dynamic-set operations takeO(lg n ) time in the worse case.

Red-black trees


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Chapter 13 P.3


13.1 Properties of red-black treeAred-black tree is a binary search treewith one extra bit of storage per node:

its color, which can either RED or BLACK.By constraining the way nodes can becolored on any path from the root to aleaf, red-black trees ensure that no

such path is more then twice as long asany other, so that three isapproximately balanced.


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Chapter 13 P.4


Each node of the tree now contains the fieldscolor, key, left, right, and p. If a child or the

parent of a node does not exist, thecorresponding pointer field of the nodecontains the value NIL. We shall regard theseNILs as being pointers to external

nodes(leaves) of the binary search tree andthe normal, key-bearing nodes as beinginternal nodes of the tree.


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Chapter 13 P.5


A binary search tree is a red-black tree if itsatisfies the following red-black properties:

1. Every node is either red or black.2. The root is black.2. The root is black.

3. Every leaf (NIL) is black

4. If a node is red, then both its children are black.

5. For each node, all paths from the node todescendant leaves contain the same number ofblack nodes.


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Chapter 13 P.6


A red-black tree with black nodes

and red nodes shaded.


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8/32Chapter 13 P.8



9/32Chapter 13 P.9


Lemma 13.1A red-black tree with n internal nodes

has height at most 2lg(n+1).


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Proof. We start by showing that the subtree

rooted at any nodex contains at least

2bh(x) 1 internal nodes. We prove thisclaim by induction on the height ofx. ifthe eight ofx is 0, thenx must be a

leaf(NIL[T]), and the sub tree rooted atx indeed contains at least 2bh(x) 1 = 20 1 =0 internal nodes.




For the inductive step, consider a node x that

has positive height and is an internal nodewith two children. Each child has a black-height of either bh(x) or bh(x) 1, dependingon whether its color is red or black,

respectively. Since the height of a child ofx isless than the height ofx itself, we can applythe inductive hypothesis to conclude thateach child has at least 2bh(x)-1 -1 internal nodes.

Thus, the subtree rooted at x contains atleast (2bh(x)-1 1)+(2bh(x)-1 1) +1 = 2bh(x) 1internal nodes, which proves the claim.




To complete the proof of the lemma, let h bethe height of the tree. According to property4, at least half the nodes on any simply path

from the root to a leaf, not including the root,must be black. Consequently, the black-heightof the root must be at least h/2; thus,

n 2h/2

-1.Moving the 1 to the left-hand side and takinglogarithms on both sides yields lg(n + 1) h/2,orh 2lg (n+1).




13.2 Rotations


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Chapter 13 P.15


An example of LEFT-ROTATE(T,x)


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Chapter 13 P.16


13.3 InsertionRB-INSERT(T,x)

1 y nil[T]

2 x root[T]3 whilex nil[T]

4 doy x

5 if key[z] < key[x]

6 thenx left[x]

7 else x right[x]

8 p[z] y


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Chapter 13 P.17


9 ify = nil[T]

10 thenroot[T] z

11 else ifkey[z] < key[y]12 thenleft[y] z

13 else right[y] z

14 left[z] nil[T]

15 right[z] nil[T]

16 color[z] RED

17 RB-INSERT-FIXUP(T,z)


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Chapter 13 P.18


RB-INSERT-FIXUP(T,z)

1 whilecolor[p[z]] = RED

2 do ifp[z] = left[p[p[z]]]3 theny right[p[p[z]]]

4 if color[y] = RED

5 thencolor[p[z]] BLACK Case 1

6 color[y] BLACK Case 1

7 color[p[p[z]]] RED Case 1

8 zp[p[z]] Case 1


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Chapter 13 P.19


9 elseifz= right[p[z]]

10 thenzp[p[z]] Case 2

11 LEFT-ROTATE(T,z) Case 212 color[p[z]] BLACK Case 3

13 color[p[p[z]]] RED Case 3

14 RIGHT-ROTATE(T,p[p[z]]) Case 3

15 else (same as then clause

with right and left exchanged)

16 color[root[T]] BLACK


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Chapter 13 P.20


The operation of RB-INSERT-

FIXUP


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Chapter 13 P.21



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Chapter 13 P.22


case 1 of the RB-INSERT


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Chapter 13 P.23


case 2 and 3 of RB-INSERT


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Chapter 13 P.24


Analysis RB-INSERT take a total ofO(lg n) time.

It never performs more than two

rotations, since the while loopterminates if case 2 or case 3 executed.


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Chapter 13 P.25


13.4 Deletion RB-DELETE(T,z)

1 ifleft[z] = nil[T] orright[z] = nil[T]

2 theny z3 elsey TREE-SUCCESSOR(z)

4 ifleft[y] nil[T]

5 thenx left[y]

6 elsex right[y]

7 p[x] p[y]

8 ifp[y] = nil[T]


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Chapter 13 P.26


9 thenroot[T] x

10 else ify = left[p[y]]

11 thenleft[p[y]] x

12 else right[p[y]] x

13 ify z

14 thenkey[z] key[y]

15 copy ys satellite data into z

16 if color[y] = BLACK

17 then RB-DELETE-FIXUP(T,x)

18 returny


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Chapter 13 P.27


RB-DELETE-FIXUP(T, x) RB-DELETE-FIXUP

1 whilex root[T] and color[x] = BLACK

2 do ifx = left[p[x]]3 thenw right[p[x]]

4 ifcolor[w] = RED

5 thencolor[w] BLACK Case1

6 color[p[x]] = RED Case1

7 LEFT-ROTATE(T,p[x]) Case1

8 w right[p[x]] Case1


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Chapter 13 P.28


9 ifcolor[right[w]] = BLACKand

color[right[w]= BLACK

10 thencolor[w] RED Case2

11 x p[x] Case2

12 elseifcolor[right[w]] = BLACK

13 thencolor[left[w]] BLACKCase3

14 color[w] RED Case3

15 RIGHT-ROTATE(T,w) Case3

16 w right[p[x]] Case3

17 color[w] color[p[x]] Case4


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Chapter 13 P.29


18 color[p[x]] BLACK Case4

19 color[right[w]] BLACK Case4

20 LEFT-ROTATE(T,p[x]) Case421 x root[T] Case4

22 else (same as then clause with right

and left exchanged)

23 color[x] BLACK


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Chapter 13 P.30


the case in the while loop of

RB-DELETE-FIXUP

C h b


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Chapter 13 P.31


C t Th L b


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Ch t 13 P 32


Analysis The RB-DELETE-FIXUP takes O(lg n) time

and performs at most three rotations.

The overall time for RB-DELETE is

therefore also O(lg n)

13 Red-Black Trees

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