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  • These slides are very well explained and very helpful. It is one of the best link I found on the internet for the deletion in red black tree.
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  • If uncles are absent that means you have to consider a NULL node there,which is black So,it does consider that cases too!
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  • THE SLIDES ARE REALLY HELPFUL BUT IN BOTTOM UP INSERTIO THE CASES WHERE UNCLES DOESN'T EXIST ARE NOT TAKEN INTO CONSIDERATION....
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  • 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
  • 5.
  • 6.
  • 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