342, TCSS Winter 2006 Lecture Notes
Balanced Binary Search Trees
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Learn about balanced binary search trees
AVL trees red-black trees
Objectives
Talk about why they work Talk about their run-time performance.
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AVL Trees
In 1960 two Russian mathematicians Georgii Maksimovich Adel'son-Vel'skii Evgenii Mikhailovich Landis developed a technique for keeping a binary search tree balanced as items are inserted into it. called AVL trees.
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Balanced AVL Tree
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extremely unbalanced tree
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AVL Trees
The balance factor of node x = height of x's right subtree minus height of x's left subtree
define height of empty tree to be -1.
binary search tree is an AVL tree if
balance factor of each node is 0, 1 or -1
Are AVL trees always completely balanced?
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Which are AVL Trees?
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Examples of AVL Trees
1 0 -1 0 0 -1 0
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0 -1 1 0
-1 0
0 0
Not AVL Trees
-1 -2 -1 0 0
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-2 2 -1 0
-1 0
2 1 0
AVL Tree Data Structure
extension of binary search tree where trees are always AVL trees. Maintain AVL property using rotations Rotations occur when tree becomes unbalanced from insertion or deletion At each node, keep track of height of subtree rooted at node (for balance factor computations)
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balancing a tree with a right rotation
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A right rotation in an AVL tree
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FIGURE 13.11 Unbalanced tree and balanced tree after a left rotation
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figure 10.12 A right-left rotation
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figure 10.13 A left-right rotation
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AVL insertion
After normal BST insert, update heights from new leaf up towards root. If balance factor changes to +2 or -2, then use rotation(s) to rebalance. Let A be the first unbalanced node found. (This is the current lowest depth node that is a non-AVL subtree). Four cases:
1. A has balance factor -2 and A's left child has balance factor 1. Then perform right rotation around A. (right rotation) 2. A has and balance factor -2 and A's left child has balance factor 1. Then perform left rotation around A's left child, followed by right rotation around A. (left-right rotation)
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AVL insertion (2)
Let A be the unbalanced node. Remaining 2 cases are mirror image of the two before.
3. A has balance factor 2 and A's right child has balance factor 1. Then perform right rotation around A's right child, followed by left rotation around A. (right-left rotation) 4. A has balance factor 2 and A's right child has balance factor 1. Then perform left rotation around A. (left rotation)
After rebalancing, continue up the tree updating heights. (and checking for imbalances in balance factor) Are all cases handled? What if new item added was A's left child or right child? What if A's child has balance factor 0?
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Right rotation to fix Case 1
right rotation (clockwise): left child becomes parent; original parent demoted to right
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Left rotation to fix Case 4
left rotation (counterclockwise): right child becomes parent; original parent demoted to left
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a single right rotation does not fix Case 2! a single left rotation also does not fix Case 3
Problem: Cases 2, 3
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Left-right rotation to fix Case 2
leftright double rotation: a left rotation of the left child, followed by a right rotation at the parent
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Right-left rotation to fix Case 3
rightleft double rotation: a right rotation of the right child, followed by a left rotation at the parent
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AVL rotation notes
right-left and left-right rotation sometimes called double rotation If subtree rooted at A was an AVL tree before the insertion, then
subtree rooted at A is always an AVL tree after applying the rotation(s) specified for each case. [Correctness] no further rotations will be made for ancestors of A. If A has balance factor 2, then
A's left child cannot have balance factor 0. A's left child has balance factor is 1 if and only if new node was added to subtree rooted at A's leftmost grandchild. A's left child has balance factor is 1 if and only if new node was added to subtree rooted at A's left child's right child. new node must be added at grandchild level of A or deeper.
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a right-left rotation after removal
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AVL deletion
Perform normal BST remove (with replacement of node to be removed with its successor). Then update heights from replacement node location upwards towards root. If balance factor changes to +2 or -2, then use rotation(s) to rebalance. Let A be the first unbalanced node found. (This is the current lowest depth node that is a non-AVL subtree). At least four cases:
1. A has balance factor -2 and A's left child has balance factor 1. Then perform right rotation around A. (right rotation) 2. A has and balance factor -2 and A's left child has balance factor 1. Then perform left rotation around A's left child, followed by right rotation around A. (left-right rotation)
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AVL deletion (2)
Let A be the unbalanced node. Additional 2 mirror image cases:
3. A has balance factor 2 and A's right child has balance factor 1. Then perform right rotation around A's right child, followed by left rotation around A. (right-left rotation) 4. A has balance factor 2 and A's right child has balance factor 1. Then perform left rotation around A. (left rotation) 5. Additional cases?
After rebalancing, continue up the tree updating heights. Must continue checking for imbalances in balance factor, and rebalancing if necessary. Are all cases handled? What if item removed was A's left child or right child? What if A's child has balance factor 0?
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Red-Black Trees
Root property: Root is BLACK. External Property: Every external node is BLACK (external nodes: null nodes)
external nodes not drawn in pictures..
Internal property: Children of a RED node are BLACK. Depth property: All external nodes have the same BLACK depth.
(BLACK depth = depth counting just black nodes)
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A tree. RedBlack Black depth 3.
30 15 10 5 40
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70 20 50 55
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60 65 80
85 90
figure 10.16 Valid red/black trees
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red/black tree after insertion
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red/black tree after removal
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RedBlack
Insertion
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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. Fix red-red violation on current level, then move upwards and repeat
Let: z be the current red node with a red parent v. (z may have been just inserted). Let u be its grandparent. u must be black. Two cases, discussed next
1. Color root black.
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Red child, red parent. Parent has a black sibling
a b v z Vl
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Fixing Tree, Case one
u w
Zl
Zr
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Case I: Rotate and recolor
Z-middle key. Black height does not change! No more red-red.
a b v Zr z u w
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Vl
Zl
Still Case One
a b v Vr z Zr
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u w
Zl
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Case I: Rotate and recolor
v-middle key
b z a v u w
Zl
Zr
Vr
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Case II
Red child, red parent. Parent has a red sibling.
a b v z Vl
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u w
Zr
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Case II: Recolor
Red-red may move up...
a b v z Vl
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u w
Zl
Zr
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Red Black Trees, Insertion
4. Fix red-red violation on current level, then move upwards (by two levels) and repeat
Let: z, v, u be current red node, red parent, and black grandparent, Case one: v has black sibling. Then rotate middle key of (z,v,u) up to the grandparent position. Color middle key (in former grandparent position) black, and its children red. Case two: v has red sibling. Then recolor grandparent red and its children black.
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Red Black Tree
Insert 10 root
10
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Red Black Tree
Insert 10 root (external nodes not shown)
10
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Red Black Tree
Insert 85
10 85
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Red Black Tree
Insert 15
10 85 15
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Red Black Tree
Rotate Change colors
15 10 85
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Red Black Tree
Insert 70
15 10 70 85
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Red Black Tree
Change Color
15 10 70 85
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Red Black Tree
Insert 20 (sibling of parent is black)
15 10 70 20 85
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Red Black Tree
Rotate
15 10 20 70 85
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Red Black Tree
Insert 60 (sibling of parent is red)
15 10 20 60
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70 85
Red Black Tree
Change Color
15 10 20 60
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70 85
Red Black Tree
Insert 30 (sibling of parent is black)
15 10 20 60 30
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70 85
Red Black Tree
Rotate
15 10 30 20 60
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70 85
Red Black Tree
Insert 50 (sibling ?)
15 10 30 20 50
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70 85 60
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Red Black Tree
Insert 50 (sibling of 70 is black!)
15 10 Child 30 20 Oops, red-red. ROTATE!
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gramps 15 70 30 60 50
55
Parent 70 85
Red Black Tree
Double Rotate Adjust colors
30 15 10 Child-Parent-Gramps Middle goes to "top" Previous top becomes child. Its right
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70 20 50 60 85
Red Black Tree
Insert 65
30 15 10 20 50 60 65
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70 85
Red Black Tree
Insert 80
30 15 10 20 50 60 65 80
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70 85
Red Black Tree
Insert 90
30 15 10 20 50 60 65 80 70 85 90
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Red Black Tree
Insert 40
30 15 10 20 50 40
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60
Red Black Tree
Adjust color
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Red Black Tree
Adjust color
30 15 10 20 50 40
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Red Black Tree
Insert 5
30 15 10 5 40
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Red Black Tree
Insert 55
30 15 10 5 40
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Delete
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