Oct 26, 2001 CSE 373, Autumn 2001 1

A Forest of Trees

• Binary search trees: simple.– good on average: O(log n)– bad in the worst case: O(n)

• AVL trees: more complicated.– good worst case bound: O(log n)

• Splay trees: “medium” complexity of code.– guaranteed good performance

over a long sequence of operations: O(m log n) over m operations (amortized running time).

Oct 26, 2001 CSE 373, Autumn 2001 2

Self-adjusting structures

• To avoid repeated accesses to deep nodes, the (tree) structure needs to alter itself after each operation.

• Idea: Since AVL rotation seems to reduce the depth of nodes, do an AVL rotate (from node to root) any time we do an operation.

Oct 26, 2001 CSE 373, Autumn 2001 3

Idea does not work

5

4

6

1

7

3

2

find(1) 5

4

6

1

7

3

2

find(2), find(3), find(4), …

Oct 26, 2001 CSE 373, Autumn 2001 4

Splay operation

• Splay(K):

• Search for K in the usual way.

• Let X be the last node inspected.– If K is in the tree, then K is in X.– If K is not in the tree, then P has an

empty child where the search for K terminated.

• Follow the path from X to the root, carrying out rotations along the way.

Oct 26, 2001 CSE 373, Autumn 2001 5

Case 0: X is the root

• Do nothing!

X

A

Oct 26, 2001 CSE 373, Autumn 2001 6

Case 1: no grandparent

• X has no grandparent (i.e., X’s parent is the root). Perform a single rotation on X and X’s parent.

P

CB

X

A

X

CB

P

A

Oct 26, 2001 CSE 373, Autumn 2001 7

Case 2: zig-zig

• X and X’s parent are both left children or both right children. Perform two single rotations:– X’s grandparent and X’s parent.– X and X’s parent.

P

CB

X

A

G

D

P

C

B

GA

X

D

Oct 26, 2001 CSE 373, Autumn 2001 8

Case 3: zig-zag

• One of X and X’s parent is a left child and the other is a right child. Perform a double rotation.

P

CB

X

A

G

D

P

CB

G

A

X

D

Oct 26, 2001 CSE 373, Autumn 2001 9

Properties of Splay(K)

• If K is in the tree, then the resulting BST will have K at the root.

• If K is not in the tree, then the root contains a key that would be the successor or predecessor of K, if K were in the tree. (What determines whether the successor predecessor is at the root?)

Oct 26, 2001 CSE 373, Autumn 2001 10

Splay trees: find, insert• Find(K): Do Splay(K). Examine

the root to see if it has K.

• Insert(I, K): Do Splay(K). If K is at the root, replace its element with I. Otherwise, create a new node from K and I and break one link to make this node the new root.

T2

J

T1

K

T2

J

T1

T

Oct 26, 2001 CSE 373, Autumn 2001 11

Splay tree: remove

• Remove(K): Do Splay(K). If the root does not contain K, do nothing. Otherwise, delete the root and Concat the two subtrees of the root.

T2

K

T1TT2T1

Splay(findMax(T1))

T2

M