The document discusses red-black trees, which are self-balancing binary search trees. It defines properties of red-black trees, including that every node must be red or black, and that every path from the root to a leaf must contain the same number of black nodes. It presents four theorems about red-black trees relating to their structure and balancing. It then describes algorithms for inserting and deleting nodes from red-black trees in a way that maintains the red-black tree properties.

1. Red-Black Trees by Thomas A. Anastasio

2. A Red-Black Tree with NULLs shown Black-Height of the tree = 4

3. Red-Black Trees Definition: A red-black tree is a binary search tree where: Every node is either red or black. Each NULL pointer is considered to be a black node If a node is red, then both of its children are black. Every path from a node to a leaf contains the same number of black nodes. Definition: The black-height of a node, n, in a red-black tree is the number of black nodes on any path to a leaf, not counting n.

4. A valid Red-Black Tree Black-Height = 2

7. Theorem 1 – Any red-black tree with root x , has at least n = 2 bh(x) – 1 internal nodes, where bh(x) is the black height of node x. Proof: by induction on height of x.

8. Theorem 2 – In a red-black tree, at least half the nodes on any path from the root to a leaf must be black. Proof – If there is a red node on the path, there must be a corresponding black node.

9. Theorem 3 – In a red-black tree, no path from any node, N, to a leaf is more than twice as long as any other path from N to any other leaf. Proof: By definition, every path from a node to any leaf contains the same number of black nodes. By Theorem 2, a least ½ the nodes on any such path are black. Therefore, there can no more than twice as many nodes on any path from N to a leaf as on any other path. Therefore the length of every path is no more than twice as long as any other path

10. Theorem 4 – A red-black tree with n internal nodes has height h <= 2 lg( n + 1). Proof: Let h be the height of the red-black tree with root x. By Theorem 2, bh(x) >= h/2 From Theorem 1, n >= 2 bh(x) - 1 Therefore n >= 2 h/2 – 1 n + 1 >= 2 h/2 lg(n + 1) >= h/2 2lg(n + 1) >= h

11. Bottom –Up Insertion Insert node as usual in BST Color the Node RED What Red-Black property may be violated? Every node is Red or Black Leaf nodes are Black NULLS If node is Red, both children must be Black Every path from node to descendant leaf must contain the same number of Blacks

12. Bottom Up Insertion Insert node; Color it RED; X is pointer to it Cases 0: X is the root -- color it black 1: Both parent and uncle are red -- color parent and uncle black, color grandparent red, point X to grandparent, check new situation 2 (zig-zag): Parent is red, but uncle is black. X and its parent are opposite type children -- color grandparent red, color X black, rotate left on parent, rotate right on grandparent 3 (zig-zig): Parent is red, but uncle is black. X and its parent are both left or both right children -- color parent black, color grandparent red, rotate right on grandparent

13. X P G U P G U Case 1 – U is Red Just Recolor and move up X

14. X P G U S X P G S U Case 2 – Zig-Zag Double Rotate X around P; X around G Recolor G and X

15. X P G U S P X G S U Case 3 – Zig-Zig Single Rotate P around G Recolor P and G

16. 11 14 15 2 1 7 5 8 Black node Red node Insert 4 into this R-B Tree

17. Insertion Practice Insert the values 2, 1, 4, 5, 9, 3, 6, 7 into an initially empty Red-Black Tree

18. Asymptotic Cost of Insertion O(lg n) to descend to insertion point O(1) to do insertion O(lg n) to ascend and readjust == worst case only for case 1 Total: O(log n)

19. Red-Black Trees Bottom-Up Deletion

20. Recall “ordinary” BST Delete If vertex to be deleted is a leaf, just delete it. If vertex to be deleted has just one child, replace it with that child If vertex to be deleted has two children, replace the value of by it’s in-order predecessor’s value then delete the in-order predecessor (a recursive step)

21. Bottom-Up Deletion Do ordinary BST deletion. Eventually a “case 1” or “case 2“ will be done (leaf or just one child). If deleted node, U, is a leaf, think of deletion as replacing with the NULL pointer, V. If U had one child, V, think of deletion as replacing U with V. What can go wrong??

22. Which RB Property may be violated after deletion? If U is red? Not a problem – no RB properties violated If U is black? If U is not the root, deleting it will change the black-height along some path

23. Fixing the problem Think of V as having an “extra” unit of blackness. This extra blackness must be absorbed into the tree (by a red node), or propagated up to the root and out of the tree. There are four cases – our examples and “rules” assume that V is a left child. There are symmetric cases for V as a right child

24. Terminology The node just deleted was U The node that replaces it is V, which has an extra unit of blackness The parent of V is P The sibling of V is S Black Node Red Node Red or Black and don’t care

25. Bottom-Up Deletion Case 1 V’s sibling, S, is Red Rotate S around P and recolor S & P NOT a terminal case – One of the other cases will now apply All other cases apply when S is Black

26. Case 1 Diagram P S V+ P S V+ Rotate P V+ S Recolor

27. Bottom-Up Deletion Case 2 V’s sibling, S, is black and has two black children . Recolor S to be Red P absorbs V’s extra blackness If P is Red, we’re done If P is Black, it now has extra blackness and problem has been propagated up the tree

28. Case 2 diagram P S V+ P+ S V Recolor and absorb Either extra black absorbed by P or P now has extra blackness

29. Bottom-Up Deletion Case 3 S is black S’s RIGHT child is RED (Left child either color) Rotate S around P Swap colors of S and P, and color S’s Right child Black This is the terminal case – we’re done

30. Case 3 diagrams P S V+ P S V Rotate P S V Recolor

31. Bottom-Up Deletion Case 4 S is Black, S’s right child is Black and S’s left child is Red Rotate S’s left child around S Swap color of S and S’s left child Now in case 3

32. Case 4 Diagrams P S V+ P S V+ Rotate P S V+ Recolor

33. 65 50 80 10 60 70 90 62 Perform the following deletions, in the order specified Delete 90, Delete 80, Delete 70

34. Red Black Trees Top-Down Insertion

35. Review of Bottom-Up Insertion In B-Up insertion, “ordinary” BST insertion was used, followed by correction of the tree on the way back up to the root This is most easily done recursively Insert winds up the recursion on the way down the tree to the insertion point Fixing the tree occurs as the recursion unwinds

36. Top-Down Insertion Strategy In T-Down insertion, the corrections are done while traversing down the tree to the insertion point. When the actual insertion is done, no further corrections are needed, so no need to traverse back up the tree. So, T-Down insertion can be done iteratively which is generally faster

37. Goal of T-D Insertion Insertion is always done as a leaf (as in ordinary BST insertion) Recall from the B-Up flow chart that if the uncle of a newly inserted node is black, we restore the RB tree properties by one or two local rotations and recoloring – we do not need to make changes further up the tree

38. Goal (2) Therefore, the goal of T-D insertion is to traverse from the root to the insertion point in such a way that RB properties are maintained, and at the insertion point, the uncle is Black. That way we may have to rotate and recolor, but not propagate back up the tree

39. Possible insertion configurations X (Red or Black) Y Z If a new node is inserted as a child of Y or Z, there is no problem since the new node’s parent is black

40. Possible insertion configurations X Y Z If new node is child of Z, no problem since Z is black. If new node is child of Y, no problem since the new node’s uncle (Z) is black – do a few rotations and recolor…. done

41. Possible insertion configurations X Y Z If new node is inserted as child of Y or Z, it’s uncle will be red and we will have to go back up the tree. This is the only case we need to avoid.

42. Top-Down Traversal X Y Z As we traverse down the tree and encounter this case, we recolor and possible do some rotations. There are 3 cases. Remember the goal – to create an insertion point at which the parent of the new node is Black, or the uncle of the new node is black.

43. Case 1 – X’s Parent is Black X Z Y P X Z P Just recolor and continue down the tree Y

44. Case 2 X’s Parent is Red (so Grandparent is Black) and X and P are both left/right children Rotate P around G Color P black Color G red Note that X’s uncle, U, must be black because it (a) was initially black, or (b) would have been made black when we encountered G (which would have had two red children -- X’s Parent and X’s uncle)

45. Case 2 diagrams X Z Y P G U S X Z Y P G U S Rotate P around G. Recolor X, Y, Z, P and G

46. Case 3 X’s Parent is Red (so Grandparent is Black) and X and P are opposite children Rotate P around G Color P black Color G red Again note that X’s uncle, U, must be black because it (a) was initially black, or (b) would have been made black when we encountered G (which would have had two red children -- X’s Parent and X’s uncle)

47. Case 3 Diagrams (1 of 2) X Z Y P G U S X Y S P G U Z Step 1 – recolor X, Y and Z. Rotate X around P.

48. Case 3 Diagrams (2 of 2) X Y S P G U Z P Y S X G U Z Step 2 – Rotate X around G. Recolor X and G

49. An exercise – insert F D T W Z V L J P E K

50. Top-Down Insert Summary P X Y Z Case 1 P is Black Just Recolor P X Y Z Case 2 P is Red X & P both left/right P X Y Z G P X Y Z G Recolor X,Y,Z P X Y Z G Rotate P around G Recolor P,G Case 3 P is Red X and P are opposite children P X Y Z G Recolor X,Y,Z Rotate X around P X P Y Z G Rotate X around G Recolor X, G Recolor X,Y,Z X P Y Z G

51. Red Black Trees Top-Down Deletion

52. Recall the rules for BST deletion If vertex to be deleted is a leaf, just delete it. If vertex to be deleted has just one child, replace it with that child If vertex to be deleted has two children, replace the value of by it’s in-order predecessor’s value then delete the in-order predecessor (a recursive step)

53. What can go wrong? If the delete node is red? Not a problem – no RB properties violated If the deleted node is black? If the node is not the root, deleting it will change the black-height along some path

54. The goal of T-D Deletion To delete a red leaf How do we ensure that’s what happens? As we traverse the tree looking for the leaf to delete, we change every node we encounter to red. If this causes a violation of the RB properties, we fix it

55. Bottom-Up vs. Top-Down Bottom-Up is recursive BST deletion going down the tree (winding up the recursion) Fixing the RB properties coming back up the tree (unwinding the recursion) Top-Down is iterative Restructure the tree on the way down so we don’t have to go back up

56. Terminology Matching Weiss text section 12.2 X is the node being examined T is X’s sibling P is X’s (and T’s) parent R is T’s right child L is T’s left child This discussion assumes X is the left child of P. As usual, there are left-right symmetric cases.

57. Basic Strategy As we traverse the tree, we change every node we visit, X, to Red. When we change X to Red, we know P is also Red (we just came from there) T is black (since P is Red, it’s children are Black)

58. Step 1 – Examine the root If both of the root’s children are Black Make the root Red Move X to the appropriate child of the root Proceed to step 2 Otherwise designate the root as X and proceed to step 2B.

59. Step 2 – the main case As we traverse down the tree, we continually encounter this situation until we reach the node to be deleted X is Black, P is Red, T is Black We are going to color X Red, then recolor other nodes and possibly do rotation(s) based on the color of X’s and T’s children 2A. X has 2 Black children 2B. X has at least one Red child

60. Case 2A X has two Black Children P T X 2A1. T has 2 Black Children 2A2. T’s left child is Red 2A3. T’s right child is Red ** if both of T’s children are Red, we can do either 2A2 or 2A3

61. Case 2A1 X and T have 2 Black Children P T X P T X Just recolor X, P and T and move down the tree

62. Case 2A2 P T X L X has 2 Black Children and T’s Left Child is Red Rotate L around T, then L around P Recolor X and P then continue down the tree L1 L2 P T X L L1 L2

63. Case 2A3 P T X X has 2 Black Children and T’s Right Child is Red Rotate T around P Recolor X, P, T and R then continue down the tree R1 R2 P R X T R2 R1 R L L

64. Case 2B X has at least one Red child Continue down the tree to the next level If the new X is Red, continue down again If the new X is Black (T is Red, P is Black) Rotate T around P Recolor P and T Back to main case – step 2

65. Case 2B Diagram P X T Move down the tree. P X T P T X If move to Black child (2B2) Rotate T around P; Recolor P and T Back to step 2, the main case If move to the Red child (2B1) Move down again

66. Step 3 Eventually, find the node to be deleted – a leaf or a node with one non-null child that is a leaf. Delete the appropriate node as a Red leaf Step 4 Color the Root Black

67. Example 1 Delete 10 from this RB Tree 15 17 16 20 23 18 13 10 7 12 6 3 Step 1 – Root has 2 Black children. Color Root Red Descend the tree, moving X to 6

68. Example 1 (cont’d) 15 17 16 20 23 18 13 10 7 12 6 3 One of X’s children is Red (case 2B). Descend down the tree, arriving at 12. Since the new X (12) is also Red (2B1), continue down the tree, arriving at 10. X

69. Example 1 (cont’d) 15 17 16 20 23 18 13 10 7 12 6 3 Step 3 -Since 10 is the node to be deleted, replace it’s value with the value of it’s only child (7) and delete 7’s red node X

70. Example 1 (cont’d) 15 17 16 20 23 18 13 7 12 6 3 The final tree after 7 has replaced 10 and 7’s red node deleted and (step 4) the root has been colored Black.

71. Example 2 Delete 10 from this RB Tree 15 17 16 20 13 10 12 6 3 4 2 Step 1 – the root does not have 2 Black children. Color the root red, Set X = root and proceed to step 2

72. Example 2 (cont’d) 15 17 16 20 13 10 12 6 3 4 2 X X has at least one Red child (case 2B). Proceed down the tree, arriving at 6. Since 6 is also Red (case 2B1), continue down the tree, arriving at 12.

73. Example 2 (cont’d) 15 17 16 20 13 10 12 6 3 4 2 X X has 2 Black children. X’s sibling (3) also has 2 black children. Case 2A1– recolor X, P, and T and continue down the tree, arriving at 10. P T

74. Example 2 (cont’d) 15 17 16 20 13 10 12 6 3 4 2 P X T X is now the leaf to be deleted, but it’s Black, so back to step 2. X has 2 Black children and T has 2 Black children – case 2A1 Recolor X, P and T. Step 3 -- Now delete 10 as a red leaf. Step 4 -- Recolor the root black

75. Example 2 Solution 15 17 16 20 13 12 6 3 4 2

76. Example 3 Delete 11 from this RB Tree 15 13 11 12 10 5 7 3 6 9 2 4 Valid and unaffected Right subtree Step 1 – root has 2 Black children. Color Root red. Set X to appropriate child of root (10)

77. Example 3 (cont’d) 15 13 11 12 10 5 7 3 6 9 2 4 X X has one Red child (case 2B) Traverse down the tree, arriving at 12.

78. Example 3 (cont’d) 15 13 11 12 10 5 7 3 6 9 4 X Since we arrived at a black node (case 2B2) assuring T is red and P is black), rotate T around P, recolor T and P Back to step 2 P T 2

79. Example 3 (cont’d) 15 13 11 12 10 5 7 3 6 9 4 X P T 2 Now X is Black with Red parent and Black sibling. X and T both have 2 Black children (case 2A1) Just recolor X, P and T and continue traversal

80. Example 3 (cont’d) 15 13 11 12 10 5 7 3 6 9 4 X P T 2 Having traversed down the tree, we arrive at 11, the leaf to be deleted, but it’s Black, so back to step 2. X and T both have two Black children. Recolor X, P and T. Step 3 -- delete 11 as a red leaf. Step 4 -- Recolor the root black

81. Example 3 Solution 13 12 10 5 7 3 6 9 4 2 15