浙江大学高级数据结构课件

1 1+ 5 ⇒ nh ≈ 2 5
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h+ 2
5
2

−1
⇒ h = O(ln n)
§5 Splay Trees Target : Any M consecutive tree operations
starting from an empty tree take at most O(M log N) time.
CHAPTER 4
Trees
§4 AVL Trees Target : Speed up searching (with insertion and
deletion)
root small large
Tool : Binary search trees
Problem : Although Tp = O( height ), but the
Adelson-VelskiiAdelson-Velskii-Landis (AVL) Trees (1962)
【Definition】An empty binary tree is height balanced. If T is a 】 nonempty binary tree with TL and TR as its left and right subtrees, then T is height balanced iff (1) TL and TR are height balanced, and of an empty tree The height (2) | hL − hR | ≤ 1 where hL and hR defined to be –1. of TL and TR , is are the heights
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§5 Splay Trees
Deletions: Step 1: Find X ;
X will be at the root.
Home Step 2: Remove X ;
There will be two subtrees work: TL and TR .
p.136 4.23 The largest element Access) a splay tree Step 3: FindMax ( TL ; will be the root of T ,
respectively.
【Definition】The balance factor BF( node ) = hL − hR . In an 】 AVL tree, BF( node ) = −1, 0, or 1. 4 3 2 1
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5 5 6 7 1 3 2 4 7 8 1 3 2 4
7 8 5
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Jan
1 00 −1
§4 AVL Trees −11 02 −
Double Rotation
0 0
May Nov
1 0 00 1 −1
−11 00 − 2
Mar
0
Aug
0
0 −1
Aug
0
May Jan
0
Apr
0
Mar LR Rotation LR Insertion AR CR BL CL OR
Sort: List the elements in increasing order Solution: inorder traversal. Get Height: Compute the heights of the nodes Solution: postorder traversal. Get Depth: Compute the depths of the nodes Solution: preorder traversal.
P A B X A C G P X A B C D Single rotation A B C X P G D B C D G Double rotation D P G X
Rotate X and P
Zig-zig
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§5 Splay Trees
Splaying not only moves the accessed node to the root, but also roughly halves k the depth of most nodes on the path. 5
0 0 00 −1 −1 1 −1 Jan
Note: Several bf’s might be changed even if p.136 4.22 we don’t need to reconstruct Double Rotation the tree.
0 − Feb July May1 Nov Feb Home work: Nov July 0 00 −1 0 −1 0 Feb July 0 Nov −1 0 June Oct June 0 p.136 4.16 May 0 June Oct 0 Create an AVL Tree Sept
height can be as bad as O( N ).
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§4 AVL Trees
〖Example〗 2 binary search trees obtained for the months 〗 of the year
Jan Feb Apr Aug Dec Entered from Jan to Dec July Feb Aug Apr Dec Jan June A balanced tree
0 or 1
0
B
C
−1 or 0
AL CL
C
CR BR
C
CL OR CR BR AL
A
CL OR CR
B
BR
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June
Oct
Sept
−0 1 −1−11 0 0 − 2
§4 AVL Trees
Apr
0 Apr Apr
0 0
Aug
1 Aug Aug
0 1 1
Jan Mar Jan 10 0 −1 −11 −11 −2 00 −2 10 −1 −1 0 0 −11 −2 Dec Mar May Dec Dec Mar
Discussion 2: How shall we move the node? Why does your method work?
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§5 Splay Trees
k1 k2 A k4 k4
k5 k4 k5 E k E4 E
k5 k5
k5 F F F
A k2 A A A
k3 k1 k3 k4 B k1 k2 k2 D k1 k3 k2 k3D k1 k3 B C C D B CC B C B
−1 2
Apr
Nov
Jan In general:
1 0
A
0
LR Rotation
0 or 1 0
A
±1
B
B
C
−1 or 0
C
BL CL
C
CR
AR BL
B
CL OR CR
A
AR
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Dec
−1 0 2 −1
July
−21 0 − 2 1
Feb
0 −1
§4 AVL Trees
Mar
0 1
RL Rotation
−1
A
AL BL
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RR
0
−2
0
A
AL
B
BR
Insertion
−1
RR Rotation BR AL
0
B
BR BL
B
BL
A
Aug
Apr
−11 02 −
§4 AVL Trees
May
2 1 0 0 0
Mar
1 0
Nov
Aug
0
Apr
Discussion 1: What can we do now?
k4 k3 1 k2 AA B k1 B C Ck3 C D D k3 k4 D E k5 E
F
F
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§5 Splay Trees
Insert: 1, 2, 3, 4, 5, 6, 7
7 6 5 4 3 2 1 1 2 4
Find: 1
7 6 5 1 4 2 3 3 Read the 32-node example given in Figures 4.52 – 4.60 5 2 3 6 4 5 7 1 6 7
§4 AVL Trees
〖Example〗 Input the months 〗
−1 0 2
Mar
May
Nov
−11 00 − 2
Mar
0 −1
Single rotation 0 0 0
May
0 0
May Nov
Mar
Nov
The trouble maker Nov is in the right subtree’s right subtree of the trouble finder Mar. Hence it is called an RR rotation. A is not necessarily In general: the root of the tree
0 00 −2 −11
−11 0 − 2
Mar
00 −1 1
Aug
0
May Jan
0
0 −1
Dec
0
May
0 0ug Apr Feb
Jan July
Dec
0
July Feb
Nov
In general: