1. TCSS 342, Winter 2006 Lecture Notes
Balanced Binary Search Trees
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2. Objectives Learn about balanced binary search trees
AVL trees
red-black trees
Talk about why they work
Talk about their run-time performance.
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3. 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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4. Balanced AVL Tree
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5. extremely unbalanced tree
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6. 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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7. Which are AVL Trees?
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8. Examples of AVL Trees 1 -1 -1 -1 1 -1
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9. Not AVL Trees -1 -2 -2 -1 2 -1 2 1 -1
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10. 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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11. balancing a tree with a right rotation
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12. A right rotation in an AVL tree
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13. FIGURE 13.11 Unbalanced tree and balanced tree after a left rotation
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14. figure 10.12 A right-left rotation
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15. figure 10.13 A left-right rotation
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16. 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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17. 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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18. Right rotation to fix Case 1
right rotation (clockwise): left child becomes parent; original parent demoted to right
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19. Left rotation to fix Case 4
left rotation (counter-clockwise): right child becomes parent; original parent demoted to left
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20. Problem: Cases 2, 3 a single right rotation does not fix Case 2!
a single left rotation also does not fix Case 3
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21. Left-right rotation to fix Case 2
left-right double rotation: a left rotation of the left child, followed by a right rotation at the parent
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22. Right-left rotation to fix Case 3
right-left double rotation: a right rotation of the right child, followed by a left rotation at the parent
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23. 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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24. a right-left rotation after removal
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25. 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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26. 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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27. 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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28. A RedBlack tree. Black depth 3.30 15 70 20 85 10 5 60 65 50 40 90 80 55
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29. figure 10.16 Valid red/black trees
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30. red/black tree after insertion
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31. red/black tree after removal
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32. RedBlack
Insertion
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33. Red Black Trees, Insertion
Find proper external node.
Insert and color node red.
No black depth violation but may violate the red-black parent-child relationship.
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
Color root black.
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34. Fixing Tree, Case one
Red child, red parent. Parent has a black sibling a b u w v z Vl Zr Zl
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35. Case I: Rotate and recolor
Z-middle key. Black height does not change! No more red-red. a b z u v w Zr Zl Vl
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36. Still Case One a b u w v Vr z Zr Zl
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37. Case I: Rotate and recolor
v-middle key a b v u z w Zl Zr Vr
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38. Case II Red child, red parent. Parent has a red sibling. a b u w v zVl Zr
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39. Case II: Recolor Red-red may move up… a b u w v z Vl Zr Zl
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40. 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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41. Red Black Tree
Insert 10 – root 10
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42. Red Black Tree Insert 10 – root (external nodes not shown) 10
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43. Red Black Tree
Insert 85 10 85
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44. Red Black Tree
Insert 15 10 85 15
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45. Red Black Tree
Rotate – Change colors 15 10 85
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46. Red Black Tree
Insert 70 15 10 85 70
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47. Red Black Tree
Change Color 15 10 85 70
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48. Red Black Tree Insert 20 (sibling of parent is black) 15 10 85 70 20
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49. Red Black Tree
Rotate 15 10 70 85 20
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50. Red Black Tree Insert 60 (sibling of parent is red) 15 10 70 85 20 60
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51. Red Black Tree
Change Color 15 10 70 85 20 60
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52. Red Black Tree Insert 30 (sibling of parent is black) 15 10 70 85 2060 30
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53. Red Black Tree
Rotate 15 10 70 85 30 20 60
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54. Red Black Tree Insert 50 (sibling ?) 15 10 70 85 30 20 60 50
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55. Red Black Tree Insert 50 (sibling of 70 is black!) gramps 15 15
 Parent 70 10 70 85 30
Child  30 20 60
Oops, red-red. ROTATE! 50
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56. Red Black Tree Double Rotate – Adjust colors 30 15 70 10 20 85 60
Child-Parent-Gramps
Middle goes to “top”
Previous top becomes child. Its right child is middles left child. 50
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57. Red Black Tree
Insert 65 30 15 70 10 20 85 60 50 65
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58. Red Black Tree
Insert 80 30 15 70 10 20 85 60 80 50 65
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59. Red Black Tree
Insert 90 30 15 70 10 20 85 60 80 90 50 65
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60. Red Black Tree
Insert 40 30 15 70 10 20 85 60 80 90 50 65 40
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61. Red Black Tree Adjust color 30 15 70 10 20 85 60 80 90 50 65 40
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62. Red Black Tree Adjust color 30 15 70 10 20 85 60 80 90 50 65 40
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63. Red Black Tree
Insert 5 30 15 70 10 20 85 60 5 80 90 50 65 40
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64. Red Black Tree
Insert 55 30 15 70 10 20 85 60 5 80 90 50 65 40 55
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65. 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.
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66. Delete demo Delete 30: Swap with 40 and delete red leaf. 30 15 70 1020 85 60 5 80 90 50 65 40 55
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67. 40 15 70 10 20 85 60 5 80 90 50 65 55
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68. Inorder successor is Black
Change colors along the traverse path so that the leaf to be deleted is RED.
Delete 15. 30 15 70 10 20 85 60 5 80 90 50 65 40 55
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69. 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
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70. Both children of X and T are black: P  Black X  Red, T  Red
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71. If X is a leaf we are done. Recall: x is the inorder successor!P X T A B
If X is a leaf we are done. Recall: x is the inorder successor!
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72. P X T C1 A D B C Zig-Zag, C1 Middle key.
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 X T C1 A D B C
Zig-Zag, C1 Middle key.
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73. Note: black depth remains unchanged!X B C D A
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74. Third case P X T C1 A B D C T middle key.
B will become P’s right child. No change in depth.
Third case P X T C1 A B D C
T middle key.
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75. T P C1 X B C D A
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76. 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:
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77. P X T C1 E C B Y A B Root of C is black Otherwise, continue
X has a red child P X T C1 E C B Y A B
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78. P C1 T X E Y A C B
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79. 30
Delete 15 15 70 10 20 85 60 5 80 90 50 65 40 55
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80. 60
Delete 15 30 70 15 85 50 65 10 20 80 90 40 55 60 5
Swap (15, 20) 30 70 20 85 50 65 10 15 80 90 40 55 5
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81. Third case: (mirror image) X (15) has two black children (Nulls) 60
Delete 15 30 70 20 85 50 65 10 15 80 90 40 55 5
Third case: (mirror image) X (15) has two black children (Nulls)
Sibling has one red and one black child.
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82. 60
Delete 15 30 70 10 85 50 65 5 20 80 90 40 55
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83. References
Lewis & Chase book, Chapter 13.
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