Multi-Way search Trees

1. 2-3 Trees:

a. Nodes may contain 1 or 2 items.

b. A node with k items has k + 1 children

c. All leaves are on same level.

Example

• A 2-3 tree storing 18 items.

20 80

5 30 70 90 100

2 4 10 25 40 50 75 85 95 110 120

Updating

• Insertion: • Find the appropriate leaf. If there is only

one item, just add to leaf.• Insert(23); Insert(15)• If no room, move middle item to parent and

split remaining two items among two children.

• Insert(3);

Insertion

• Insert(3);

20 80

5 30 70 90 100

2 3 4 10 15 23 25 40 50 75 85 95 110 120

Insert(3);

• In mid air…20 80

5 30 70 90 100

2 10 15 23 25 40 50 75 85 95 110 120

3

4

Done….

20 80

3 5 30 70 90 100

2 10 15 23 25 40 50 75 85 95 110 1204

Tree grows at the root…

• Insert(45);

20 80

3 5 30 70 90 100

2 10 25 40 45 50 75 85 95 110 1204

• New root:

80

3 5 30 90 100

2 10 25 40 75 85 95 110 1204

45

20

50

70

Delete

• If item is not in a leaf exchange with in-order successor.

• If leaf has another item, remove item.

• Examples: Remove(110);

• (Insert(110); Remove(100); )

• If leaf has only one item but sibling has two items: redistribute items. Remove(80);

Remove(80);• Step 1: Exchange 80 with in-order

successor.

85

3 5 30 90 100

2 10 25 40 75 80 95 110 1204

45

20

50

70

Remove(80);• Redistribute

85

3 5 30 95 110

2 10 25 40 75 90 100 1204

45

20

50

70

Some more removals…

• Remove(70);

Swap(70, 75);

Remove(70);

“Merge” Empty node with sibling;

Join parent with node;

Now every node has k+1 children except that one node has 0 items and one child.

Sibling can spare an item: redistribute.95 110

Delete(70)

85

3 5 30 95 110

2 10 25 40 90 100 1204

45

20

50

75

New tree:• Delete(85) will “shrink” the tree.

95

3 5 30 110

2 10 25 40 90 100 1204

45

20

50

85

Details

• 1. Swap(85, 90) //inorder successor• 2. Remove(85) //empty node created• 3. Merge with sibling• 4. Drop item from parent// (50,90) empty Parent• 5. Merge empty node with sibling, drop item from

parent (95)• 6. Parent empty, merge with sibling drop item.

Parent (root) empty, remove root.

“Shorter” 2-3 Tree

3 5 30 95 110

2 10 25 40 100 1204

20 45

50 90

Deletion Summary

• If item k is present but not in a leaf, swap with inorder successor;

• Delete item k from leaf L.• If L has no items: Fix(L);• Fix(Node N);• //All nodes have k items and k+1 children• // A node with 0 items and 1 child is

possible, it will have to be fixed.

Deletion (continued)

• If N is the root, delete it and return its child as the new root.

• Example: Delete(8);

5

3 8

5

3

1 2

3 5

3

Return 3 5

Deletion (Continued)

• If a sibling S of N has 2 items distribute items among N, S and the parent P; if N is internal, move the appropriate child from S to N.

• Else bring an item from P into S;• If N is internal, make its (single) child the

child of S; remove N.• If P has no items Fix(P) (recursive call)

(2,4) Trees

• Size Property: nodes may have 1,2,3 items.

• Every node, except leaves has size+1 children.

• Depth property: all leaves have the same depth.

• Insertion: If during the search for the leaf you encounter a “full” node, split it.

(2,4) Tree

10 45 60

3 8 25 50 55 70 90 100

Insert(38);

Insert(38);

3 8 25 38 50 55 70 90 100

45

10

60

Insert(105)

• Insert(105);

3 8 25 38 50 55 100 105

45

10 60 90

70

Removal

• As with BS trees, we may place the node to be removed in a leaf.

• If the leaf v has another item, done.• If not, we have an UNDERFLOW.• If a sibling of v has 2 or 3 items, transfer an

item.• If v has 2 or 3 siblings we perform a transfer

Removal

• If v has only one sibling with a single item we drop an item from the parent to the sibling, remove v. This may create an underflow at the parent. We “percolate” up the underflow. It may reach the root in which case the root will be discarded and the tree will “shrink”.

Delete(15)

35

20 60

6 15 40 50 70 80 90

Delete(15)

35

20 60

6 40 50 70 80 90

Continued

• Drop item from parent

35

60

6 20 40 50 70 80 90

Fuse

35

60

6 20 40 50 70 80 90

Drop item from root

• Remove root, return the child.

35 60

6 20 40 50 70 80 90

Summary

• Both 2-3 trees and 2-4 trees make it very easy to maintain balance.

• Insertion and deletion easier for 2-4 tree.

• Cost is waste of space in each node. Also extra comparison inside each node.

• Does not “extend” binary trees.

Red-Black Trees

• Root property: Root is BLACK.

• External Property: Every external node is BLACK (external nodes: null nodes)

• Internal property: Children of a RED node are BLACK.

• Depth property: All external nodes have the same BLACK depth.

A RedBlack tree.Black depth 3.

30

15 70

20 8510

5

60

6550

40

9080

55

RedBlack

Insertion

Red Black Trees, Insertion

1. Find proper external node.

2. Insert and color node red.

3. No black depth violation but may violate the red-black parent-child relationship.

4. Let: z be the inserted node, v its parent and u its grandparent. If v is red then u must be black.

Color adjustments.• Red child, red parent. Parent has a black

sibling (Zig-Zag).

a

b u

v w

zVl

ZlZr

Rotation• Z-middle key. Black height does not

change! No more red-red.

a

b z

v

w

u

Vl Zl Zr

Color adjustment II

a

b u

v w

z

Vr

ZlZr

Rotation II• v-middle key a

b v

z u

ZrZlw

Vr

Recoloring

• Red child, red parent. Parent has a red sibling.

a

b u

v w

zVl

Zr

Color adjustment

• Red-red may move up…

a

b u

v w

zVl

ZrZl

Red Black Tree

• Insert 10 – root

10

Red Black Tree

• Insert 10 – root (external nodes not shown)

10

Red Black Tree

• Insert 85

10

85

Red Black Tree

• Insert 15

10

85

15

Red Black Tree

• Rotate – Change colors

15

10 85

Red Black Tree

• Insert 70

15

10 85

70

Red Black Tree

• Change Color

15

10 85

70

Red Black Tree

• Insert 20 (sibling of parent is black)

20

15

10 85

70

Red Black Tree

• Rotate

15

10 70

20 85

Red Black Tree

• Insert 60 (sibling of parent is red)

15

10 70

20 85

60

Red Black Tree

• Change Color

15

10 70

20 85

60

Red Black Tree

• Insert 30 (sibling of parent is black)

15

10 70

20 85

60

30

Red Black Tree

• Rotate

15

10 70

30 85

6020

Red Black Tree

• Insert 50 (sibling ?)

15

10 70

30 85

6020

50

Red Black Tree

• Insert 50 (sibling of 70 is black!)

15

10 70

30 85

6020

50Oops, red-red. ROTATE!

Child 30

Parent 70

gramps 15

Red Black Tree

• Double Rotate – Adjust colors

30

15 70

20 8510 60

50Child-Parent-Gramps

Middle goes to “top”

Previous top becomes child. Its right child is middles left child.

Red Black Tree

• Insert 65

30

15 70

20 8510 60

6550

Red Black Tree

• Insert 80

30

15 70

20 8510 60

6550 80

Red Black Tree

• Insert 90

30

15 70

20 8510 60

6550 9080

Red Black Tree

• Insert 40

30

15 70

20 8510 60

6550

40

9080

Red Black Tree

• Adjust color

30

15 70

20 8510 60

6550

40

9080

Red Black Tree

• Insert 5

30

15 70

20 8510

5

60

6550

40

9080

Red Black Tree

• Insert 55

30

15 70

20 8510

5

60

6550

40

9080

55

Delete

• We first note that a red node is either a leaf or must have two children.

• Also, if a black node has a single child it must be a red leaf.

• Swap X with inorder successor.• If inorder successor is red, (must be a leaf) delete.

If it is a single child parent, delete and change its child color to black. In both cases the resulting tree is a legit red-black tree.

Delete demo

• Delete 30: Swap with 40 and delete red leaf.

30

15 70

20 8510

5

60

6550

40

9080

55

40

15 70

20 8510

5

60

6550 9080

55

Inorder successor is BlackChange colors along the traverse path so that the leaf to be deleted is RED.

Delete 15.30

15 70

20 8510

5

60

6550

40

9080

55

General strategy

• As you traverse the tree to locate the inorder successor let X be the current node, T its sibling and P the parent.

• Color the root red.

• Retain: “the color of P is red.”

• If all children of X and T are black:

• P Black, X Red, T Red

X T

P

Both children of X and T are black:

P Black X Red, T Red

A B

X T

P

If X is a leaf we are done. Recall: x is the inorder successor!

A B

X T

P

C1

B C

D

Zig-Zag, C1 Middle key.

A

Even though we want to proceed with X we have a red-red violation that needs to be fixed.

T has a red child.

Even though we want to proceed with X we have a red-red violation that needs to be fixed.

T has a red child.

P T

C1

B C D

X

A

Note: black depth remains unchanged!

Note: black depth remains unchanged!

Third case

X T

P

C1

B

C D

A

T middle key.

B will become P’s right child. No change in depth.

B will become P’s right child. No change in depth.

P C1

T

B C DA

X

• If both children of T are red select one of the two rotations.

• If the right child of X is red make it the new parent (it is on the inorder-successor path).

• If the left child of X is red:

T

P

X

EC1

B

Y

C

AB

Left as a drill.

Root of C is black

Otherwise, continue

X has a red child

Root of C is black

Otherwise, continue

X has a red child

T

P

XE

C1

Y

CA

B

30

15 70

20 8510

5

60

6550

40

9080

55

Delete 15

30

20

70

10

8515

5

60

6550

40 908055

Delete 15

30

15

70

10

8520

5

60

6550

40 908055

Swap (15, 20)

30

15

70

10

8520

5

60

6550

40 908055

Delete 15

Third case: (mirror image) X (15) has two black children (Nulls)

Sibling has one red and one black child.

30 70

10 85

205

60

6550

40 908055

Delete 15