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BST, RBT, Tree

BST, RBT, Tree

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    Red black tree Red black tree Presentation Transcript

    • Algorithms Sandeep Kumar Poonia Head Of Dept. CS/IT B.E., M.Tech., UGC-NET LM-IAENG, LM-IACSIT,LM-CSTA, LM-AIRCC, LM-SCIEI, AM-UACEE
    • Algorithms BST Red Black Tree
    • Binary search trees ● Binary search trees are an important data structure for dynamic sets. ● Accomplish many dynamic-set operations in O(h) time, where h=height of tree. ● we represent a binary tree by a linked data structure in which each node is an object. ● T:root points to the root of tree T .
    • Binary search trees ● Each node contains the attributes ■ key (and possibly other satellite data). ■ left: points to left child. ■ right: points to right child. ■ p: points to parent. T.root.p = NIL.
    • Binary search trees ● Stored keys must satisfy the binary-search- tree property. ■ If y is in left subtree of x, then y.key <= x.key. ■ If y is in right subtree of x, then y.key >= x.key.
    • Binary search trees (a) A binary search tree on 6 nodes with height 2. (b) A less efficient binary search tree with height 4 that contains the same keys.
    • Binary Search Trees A binary search tree Not a binary search tree
    • Binary search trees ● The binary-search-tree property allows us to print out all the keys in a binary search tree in sorted order by a simple recursive algorithm, called an inorder tree walk. It takes ‚ time to walk an n-node binary search tree
    • Binary search trees ● A: prints elements in sorted (increasing) order ● This is called an inorder tree walk ■ Preorder tree walk: print root, then left, then right ■ Postorder tree walk: print left, then right, then root
    • Tree traversal ● Used to print out the data in a tree in a certain order ● Pre-order traversal ■ Print the data at the root ■ Recursively print out all data in the left subtree ■ Recursively print out all data in the right subtree
    • Preorder, Postorder and Inorder ● Preorder traversal ■ node, left, right ■ prefix expression ○ ++a*bc*+*defg
    • Preorder, Postorder and Inorder ● Postorder traversal ■ left, right, node ■ postfix expression ○ abc*+de*f+g*+ ● Inorder traversal ■ left, node, right. ■ infix expression ○ a+b*c+d*e+f*g
    • Binary search trees the running time of TREE-SEARCH is O(h), where h is the height of the tree.
    • Binary search trees
    • insert ● Proceed down the tree as you would with a find ● If X is found, do nothing (or update something) ● Otherwise, insert X at the last spot on the path traversed ● Time complexity = O(height of the tree)
    • delete ● When we delete a node, we need to consider how we take care of the children of the deleted node. ■ This has to be done such that the property of the search tree is maintained.
    • delete Three cases: (1) the node is a leaf ■ Delete it immediately (2) the node has one child ■ Adjust a pointer from the parent to bypass that node
    • delete (3) the node has 2 children ■ replace the key of that node with the minimum element at the right subtree ■ delete the minimum element ○ Has either no child or only right child because if it has a left child, that left child would be smaller and would have been chosen. So invoke case 1 or 2. ● Time complexity = O(height of the tree)
    • Balanced Binary Search Trees A binary search tree can implement any of the basic dynamic-set operations in O(h) time. These operations are O(lgn) if tree is “balanced”. There has been lots of interest in developing algorithms to keep binary search trees balanced, including 1st type: insert nodes as is done in the BST insert, then rebalance tree  Red-Black trees  AVL trees  Splay trees 2nd type: allow more than one key per node of the search tree:  2-3 trees  2-3-4 trees  B-trees
    • Red-Black Trees (RBT) Red-Black tree: BST in which each node is colored red or black. Constraints on the coloring and connection of nodes ensure that no root to leaf path is more than twice as long as any other, so tree is approximately balanced. Each RBT node contains fields left, right, parent, color, and key. L E F T PARENT KEY COLOR R I G H T
    • Red-Black Trees ● Red-black trees: ■ Binary search trees augmented with node color ■ Operations designed to guarantee that the height h = O(lg n) ● First: describe the properties of red-black trees ● Then: prove that these guarantee h = O(lg n) ● Finally: describe operations on red-black trees
    • Red-Black Properties ● The red-black properties: 1. Every node is either red or black 2. Every leaf (NULL pointer) is black ○ Note: this means every “real” node has 2 children 3. If a node is red, both children are black ○ Note: can’t have 2 consecutive reds on a path 4. Every path from node to descendent leaf contains the same number of black nodes 5. The root is always black
    • Red-Black Trees ● Put example on board and verify properties: 1. 2. 3. 4. Every node is either red or black Every leaf (NULL pointer) is black If a node is red, both children are black Every path from node to descendent leaf contains the same number of black nodes 5. The root is always black ● black-height: # black nodes on path to leaf ■ Label example with h and bh values
    • Black Height bh(x) Black-height of a node x: bh(x) is the number of black nodes (including the NIL leaf) on the path from x to a leaf, not counting x itself. 2 20 2 18 1 17 0 0 1 22 1 19 0 Every node has a black-height, bh(x). For all NIL leaves, bh(x) = 0. 1 21 1 25 0 0 0 0 0 For root x, bh(x) = bh(T).
    • Height of a Red-black Tree h=4 26 bh=2 ● Example: ● Height of a node: ■ Number of edges in a longest path to a leaf. ● Black-height of a node 17 h=1 bh=1 h=2 30 bh=1 bh(x) is the number of black nodes on path from x to leaf, not counting x. nil[T] h=3 41 bh=2 h=1 bh=1 38 h=2 47 bh=1 h=1 50 bh=1
    • Height of Red-Black Trees ● What is the minimum black-height of a node ● ● ● ● with height h? A height-h node has black-height  h/2 Proof : By property 4, h/2 nodes on the path from the node to a leaf are red. Hence are black.
    • Height of Red-Black Trees ● The subtree rooted at any node x contains >= 2bh(x)_1 internal nodes. ● Proof :By induction height of x =0x is a leafbh(x)=0. The subtree rooted at x has 0 internal nodes. 20 -1 = 0. ● Let the height of x be h and bh(x) = b. ● Any child of x has height h -1 and ● black-height either b (if the child is red) or ● b -1 (if the child is black) ● By induction each child has >= 2bh(x)-1-1 internal nodes ● Thus, the subtree rooted at x contains >= 2(2bh(x)-1-1)+1 ● = 2bh(x)-1(internal Nodes)
    • Height of Red-Black Trees ● Theorem: A red-black tree with n internal nodes has height h  2 lg(n + 1) ● How do you suppose we’ll prove this? ● Proof: The subtree rooted at any node x contains ● >= 2bh(x)_1 internal nodes. ● A height-h node has black-height  h/2 n  2h/2 -1 n + 1 2h/2 lg(n+1)  h/2  h  2lg(n+1) ● Thus ● ●
    • RB Trees: Worst-Case Time ● So we’ve proved that a red-black tree has O(lg n) height ● Corollary: These operations take O(lg n) time: ■ Minimum(), Maximum() ■ Successor(), Predecessor() ■ Search() ● Insert() and Delete(): ■ Will also take O(lg n) time ■ But will need special care since they modify tree
    • Red-Black Trees: An Example ● Color this tree: 7 5 9 12 Red-black properties: 1. Every node is either red or black 2. Every leaf (NULL pointer) is black 3. If a node is red, both children are black 4. Every path from node to descendent leaf contains the same number of black nodes 5. The root is always black
    • Red-Black Trees: The Problem With Insertion ● Insert 8 ■ Where does it go? 7 5 9 12 1. 2. 3. 4. 5. Every node is either red or black Every leaf (NULL pointer) is black If a node is red, both children are black Every path from node to descendent leaf contains the same number of black nodes The root is always black
    • Red-Black Trees: The Problem With Insertion ● Insert 8 ■ Where does it go? ■ What color 7 5 should it be? 1. 2. 3. 4. 5. Every node is either red or black Every leaf (NULL pointer) is black If a node is red, both children are black Every path from node to descendent leaf contains the same number of black nodes The root is always black 9 8 12
    • Red-Black Trees: The Problem With Insertion ● Insert 8 ■ Where does it go? ■ What color 7 5 should it be? 1. 2. 3. 4. 5. Every node is either red or black Every leaf (NULL pointer) is black If a node is red, both children are black Every path from node to descendent leaf contains the same number of black nodes The root is always black 9 8 12
    • Red-Black Trees: The Problem With Insertion ● Insert 11 ■ Where does it go? 7 5 9 8 1. 2. 3. 4. 5. Every node is either red or black Every leaf (NULL pointer) is black If a node is red, both children are black Every path from node to descendent leaf contains the same number of black nodes The root is always black 12
    • Red-Black Trees: The Problem With Insertion ● Insert 11 ■ Where does it go? ■ What color? 7 5 9 8 12 11 1. 2. 3. 4. 5. Every node is either red or black Every leaf (NULL pointer) is black If a node is red, both children are black Every path from node to descendent leaf contains the same number of black nodes The root is always black
    • Red-Black Trees: The Problem With Insertion ● Insert 11 ■ Where does it go? ■ What color? ○ Can’t be red! (#3) 7 5 9 8 12 11 1. 2. 3. 4. 5. Every node is either red or black Every leaf (NULL pointer) is black If a node is red, both children are black Every path from node to descendent leaf contains the same number of black nodes The root is always black
    • Red-Black Trees: The Problem With Insertion ● Insert 11 ■ Where does it go? ■ What color? ○ Can’t be red! (#3) ○ Can’t be black! (#4) 1. 2. 3. 4. 5. 7 5 Every node is either red or black Every leaf (NULL pointer) is black If a node is red, both children are black Every path from node to descendent leaf contains the same number of black nodes The root is always black 9 8 12 11
    • Red-Black Trees: The Problem With Insertion ● Insert 11 ■ Where does it go? ■ What color? ○ Solution: recolor the tree 1. 2. 3. 4. 5. 7 5 Every node is either red or black Every leaf (NULL pointer) is black If a node is red, both children are black Every path from node to descendent leaf contains the same number of black nodes The root is always black 9 8 12 11
    • Red-Black Trees: The Problem With Insertion ● Insert 10 ■ Where does it go? 7 5 9 8 12 11 1. 2. 3. 4. 5. Every node is either red or black Every leaf (NULL pointer) is black If a node is red, both children are black Every path from node to descendent leaf contains the same number of black nodes The root is always black
    • Red-Black Trees: The Problem With Insertion ● Insert 10 ■ Where does it go? ■ What color? 7 5 9 8 12 11 1. 2. 3. 4. 5. Every node is either red or black Every leaf (NULL pointer) is black If a node is red, both children are black Every path from node to descendent leaf contains the same number of black nodes The root is always black 10
    • Red-Black Trees: The Problem With Insertion ● Insert 10 ■ Where does it go? 7 5 ■ What color? ○ A: no color! Tree is too imbalanced ○ Must change tree structure to allow recoloring ■ Goal: restructure tree in O(lg n) time 9 8 12 11 10
    • RB Trees: Rotation ● Our basic operation for changing tree structure is called rotation: y x x rightRotate(y) C A y leftRotate(x) A B B C ● Does rotation preserve inorder key ordering? ● What would the code for rightRotate() actually do?
    • RB Trees: Rotation y x A x rightRotate(y) C A B y B ● A lot of pointer manipulation ■ x keeps its left child ■ y keeps its right child ■ x’s right child becomes y’s left child ■ x’s and y’s parents change C
    • RB Trees: Rotation y x A x leftRotate(x) rightRotate(y) C A B y B ● A lot of pointer manipulation ■ x keeps its left child ■ y keeps its right child ■ y’s right child becomes x’s left child ■ x’s and y’s parents change C
    • RB Trees: Rotation Left-Rotate (T, x) 1. y  right[x] // Set y. 2. right[x]  left[y] //Turn y’s left subtree into x’s right subtree. 3. if left[y]  nil[T ] 4. then p[left[y]]  x 5. p[y]  p[x] // Link x’s parent to y. 6. if p[x] == nil[T ] 7. then root[T ]  y 8. else if x == left[p[x]] 9. then left[p[x]]  y 10. else right[p[x]]  y 11. left[y]  x // Put x on y’s left. 12. p[x]  y
    • RB Trees: Rotation right-Rotate (T, y) 1. x  right[y] 2. left[y]  right[x] 3. If right[x]  nil[T ] 4. then p[right[x]]  y 5. p[x]  p[y] 6. if p[y] == nil[T ] 7. then root[T ]  x 8. else if y == left[p[y]] 9. then left[p[y]]  x 10. else right[p[y]]  x 11. right[x]  y 12. p[y]  x
    • Rotation Example ● Rotate left about 9: 7 5 9 8 12 11
    • Rotation Example ● Rotate left about 9: 7 5 12 9 8 11
    • Red-Black Trees: Insertion ● Remember: 1. Insert nodes one at a time, and after every Insertion balance the tree. 2. Every node inserted starts as a Red node. 3. Consult the cases, Every time two Red nodes touch must rebalance at that point. 4. The root will always be Black.
    • Red-Black Trees: Insertion ● If we insert, what color to make the new node? ■ Red? Might violate property 3(If a node is red, both children are black). ■ Black? Might violate property 4(Every path from node to descendent leaf contains the same number of black nodes). Insertion: the basic idea ■ Insert x into tree, color x red ■ Only r-b property 3 might be violated (if p[x] red) ○ If so, move violation up tree until a place is found where it can be fixed ■ Total time will be O(lg n)
    • Reminder: Red-black Properties 1. Every node is either red or black 2. Every leaf (NULL pointer) is black 3. If a node is red, both children are black 4. Every path from node to descendent leaf contains the same number of black nodes 5. The root is always black
    • Insertion in RB Trees ● Insertion must preserve all red-black properties. ● Should an inserted node be colored Red? Black? ● Basic steps: ■ Use Tree-Insert from BST (slightly modified) to insert a node x into T. ○ Procedure RB-Insert(x). ■ Color the node x red. ■ Fix the modified tree by re-coloring nodes and performing rotation to preserve RB tree property. ○ Procedure RB-Insert-Fixup.
    • Algorithm: Insertion We have detected a need for balance when z is red and his parent too. • If z has a red uncle: colour the parent and uncle black, and grandparent red. z Balanced trees: Red-black trees
    • Algorithm: Insertion We have detected a need for balance when z is red and his parent too. • If z has a red uncle: colour the parent and uncle black, and grandparent red. • If z is a left child and has a black uncle: colour the parent black and the grandparent red, then rotateRight(z.parent.parent) Balanced trees: Red-black trees
    • Algorithm: Insertion We have detected a need for balance when z is red and his parent too. • If z has a red uncle: colour the parent and uncle black, and grandparent red. right child and has black uncle: colour the parent black and • If z is a left child and has a a black uncle, then rotateLeft(z.parent) andgrandparent red, then rotateRight(z.parent.parent) the Balanced trees: Red-black trees
    • Algorithm: Insertion Let’s insert 4, 7, 12, 15, 3 and 5. 4 7 1 2 Double red violation! It also shows it’s unbalanced… Balanced trees: Red-black trees
    • Algorithm: Insertion Let’s insert 4, 7, 12, 15, 3 and 5. 7 4 What should we do? Nothing, no double red. 3 12 5 15 Double red violation. We can’t have a better balance, and there is a red uncle… Balanced trees: Red-black trees
    • Insertion RB-Insert(T, z) 1. y  nil[T] 2. x  root[T] 3. while x  nil[T] 4. do y  x 5. if key[z] < key[x] 6. then x  left[x] 7. else x  right[x] 8. p[z]  y 9. if y = nil[T] 10. then root[T]  z 11. else if key[z] < key[y] 12. then left[y]  z 13. else right[y]  z RB-Insert(T, z) Contd. 14. left[z]  nil[T] 15. right[z]  nil[T] 16. color[z]  RED 17. RB-Insert-Fixup (T, z) How does it differ from the TreeInsert procedure of BSTs? Which of the RB properties might be violated? Fix the violations by calling RBInsert-Fixup.
    • Insertion – Fixup ● Which property might be violated? 1. OK(Every node is either red or black) 2. If z is the root, then there’s a violation. Otherwise, OK(The root is black) 3. OK(Every leaf (NIL) is black) 4. If z.p is red, there’s a violation: both z and z.p are red(If a node is red, then both its children are black) ● OK(For each node, all simple paths from the node to descendant leaves contain the same number of black nodes) ● Remove the violation by calling RB-INSERT-FIXUP:
    • Insertion – Fixup ● Problem: we may have one pair of consecutive reds where we did the insertion. ● Solution: rotate it up the tree and away… Three cases have to be handled…
    • Insertion – Fixup RB-Insert-Fixup (T, z) 1. while color[p[z]] == RED 2. do if p[z] == left[p[p[z]]] 3. then y  right[p[p[z]]] 4. if color[y] == RED 5. then color[p[z]]  BLACK // Case 1 6. color[y]  BLACK // Case 1 7. color[p[p[z]]]  RED // Case 1 8. z  p[p[z]] // Case 1
    • Insertion – Fixup RB-Insert-Fixup(T, z) (Contd.) 9. else if z == right[p[z]] // color[y]  RED 10. then z  p[z] // Case 2 11. LEFT-ROTATE(T, z) // Case 2 12. color[p[z]]  BLACK // Case 3 13. color[p[p[z]]]  RED // Case 3 14. RIGHT-ROTATE(T, p[p[z]]) // Case 3 15. else (if p[z] = right[p[p[z]]])(same as 10-14 16. with “right” and “left” exchanged) 17. color[root[T ]]  BLACK
    • Insertion – Fixup A node z ’after insertion. Because both z and its parent z.p are red, a violation of property 4 occurs. Since z’s uncle y is red, case 1 in the code applies. We recolor nodes and move the pointer z up the tree
    • Insertion – Fixup Once again, z and its parent are both red, but z’s uncle y is black. Since z is the right child of z.p, case 2 applies. We perform a left rotation,
    • Insertion – Fixup Now, z is the left child of its parent, and case 3 applies. Recoloring and right rotation yield the tree
    • Insertion – Fixup
    • Correctness Loop invariant: ● At the start of each iteration of the while loop, ■ z is red. ■ If p[z] is the root, then p[z] is black. ■ There is at most one red-black violation: ○ Property 2: z is a red root, or ○ Property 4: z and p[z] are both red.
    • Correctness – Contd. ● Initialization:  ● Termination: The loop terminates only if p[z] is black. Hence, property 4 is OK. The last line ensures property 2 always holds. ● Maintenance: We drop out when z is the root (since then p[z] is sentinel nil[T ], which is black). When we start the loop body, the only violation is of property 4. ■ There are 6 cases, 3 of which are symmetric to the other 3. We consider cases in which p[z] is a left child. ■ Let y be z’s uncle (p[z]’s sibling).
    • Case 1 – uncle y is red p[p[z]] new z C C p[z] y A D z  A   B   D    B z is a right child here. Similar steps if z is a left child.   ● p[p[z]] (z’s grandparent) must be black, since z and p[z] are both red and there are no other violations of property 4. ● Make p[z] and y black  now z and p[z] are not both red. But property 5 might now be violated. ● Make p[p[z]] red  restores property 5. ● The next iteration has p[p[z]] as the new z (i.e., z moves up 2 levels).
    • Case 1 – uncle y is red We take the same action whether z is a right child or z is a left child. Each subtree has a black root, and each has the same black-height. The code for case 1 changes the colors of some nodes, preserving property 5: all downward simple paths from a node to a leaf have the same number of blacks. The while loop continues with node z’s grandparent z.p.p as the new z. Any violation of property 4 can now occur only between the new z, which is red, and its parent, if it is red as well.
    • Case 2 – y is black, z is a right child C C p[z] p[z]  y A  y B z  z B    A   ● Left rotate around p[z], p[z] and z switch roles  now z is a left child, and both z and p[z] are red. ● Takes us immediately to case 3.
    • Case 3 – y is black, z is a left child B C p[z]  y B z  A  A  C    ● Make p[z] black and p[p[z]] red. ● Then right rotate on p[p[z]]. Ensures property 4 is maintained. ● No longer have 2 reds in a row. ● p[z] is now black  no more iterations. 
    • Cases 2 and 3 of the procedure RB-INSERT-FIXUP As in case 1, property 4 is violated in either case 2 or case 3 because z and its parent z.p are both red. Each subtree has a black root , and each has the same black-height. We transform case 2 into case 3 by a left rotation, which preserves property 5: all downward simple paths from a node to a leaf have the same number of blacks. Case 3 causes some color changes and a right rotation, which also preserve property 5. The while loop then terminates, because property 4 is satisfied: there are no longer two red nodes in a row.
    • Algorithm Analysis ● O(lg n) time to get through RB-Insert up to the call of RB-Insert-Fixup. ● Within RB-Insert-Fixup: ■ Each iteration takes O(1) time. ■ Each iteration but the last moves z up 2 levels. ■ O(lg n) levels  O(lg n) time. ■ Thus, insertion in a red-black tree takes O(lg n) time. ■ Note: there are at most 2 rotations overall.
    • Deletion ● Deletion, like insertion, should preserve all the RB properties. ● The properties that may be violated depends on the color of the deleted node. ■ Red – OK. Why? ■ Black? ● Steps: ■ Do regular BST deletion. ■ Fix any violations of RB properties that may result.
    • Deletion RB-Delete(T, z) 1. if left[z] == nil[T] or right[z] == nil[T] 2. then y  z 3. else y  TREE-SUCCESSOR(z) 4. if left[y] == nil[T ] 5. then x  left[y] 6. else x  right[y] 7. p[x]  p[y] // Do this, even if x is nil[T]
    • Deletion RB-Delete (T, z) (Contd.) 8. if p[y] == nil[T ] 9. then root[T ]  x 10. else if y == left[p[y]] 11. then left[p[y]]  x 12. else right[p[y]]  x 13. if y == z 14. then key[z]  key[y] 15. copy y’s satellite data into z 16. if color[y] == BLACK 17. then RB-Delete-Fixup(T, x) 18. return y The node passed to the fixup routine is the lone child of the spliced up node, or the sentinel.
    • RB Properties Violation ● If the delete node is red? Not a problem – no RB properties violated ● If y is black, we could have violations of redblack properties: ■ Prop. 1. OK. ■ Prop. 2. If y is the root and x is red, then the root has become red. ■ Prop. 3. OK. ■ Prop. 4. Violation if p[y] and x are both red. ■ Prop. 5. Any path containing y now has 1 fewer black
    • RB Properties Violation ● Prop. 5. Any path containing y now has 1 fewer black node. ■ Correct by giving x an “extra black.” ■ Add 1 to count of black nodes on paths containing x. ■ Now property 5 is OK, but property 1 is not. ■ x is either doubly black (if color[x] == BLACK) or red & black (if color[x] == RED). ■ The attribute color[x] is still either RED or BLACK. No new values for color attribute. ■ In other words, the extra blackness on a node is by virtue of x pointing to the node. ● Remove the violations by calling RB-Delete-Fixup.
    • Deletion – Fixup RB-Delete-Fixup(T, x) 1. while x  root[T ] and color[x] == BLACK 2. do if x == left[p[x]] 3. then w  right[p[x]] 4. if color[w] == RED 5. then color[w]  BLACK // Case 1 6. color[p[x]]  RED // Case 1 7. LEFT-ROTATE(T, p[x]) // Case 1 8. w  right[p[x]] // Case 1
    • RB-Delete-Fixup(T, x) (Contd.) /* x is still left[p[x]] */ 9. if color[left[w]] == BLACK and color[right[w]] == BLACK 10. then color[w]  RED // Case 2 11. x  p[x] // Case 2 12. else if color[right[w]] == BLACK 13. then color[left[w]]  BLACK // Case 3 14. color[w]  RED // Case 3 15. RIGHT-ROTATE(T,w) // Case 3 16. w  right[p[x]] // Case 3 17. color[w]  color[p[x]] // Case 4 18. color[p[x]]  BLACK // Case 4 19. color[right[w]]  BLACK // Case 4 20. LEFT-ROTATE(T, p[x]) // Case 4 21. x  root[T ] // Case 4 22. else (same as then clause with “right” and “left” exchanged) 23. color[x]  BLACK
    • Deletion – Fixup ● Idea: Move the extra black up the tree until x points to a red & black node  turn it into a black node, ● x points to the root  just remove the extra black, or ● We can do certain rotations and recolorings and finish. ● Within the while loop: ■ x always points to a nonroot doubly black node. ■ w is x’s sibling. ■ w cannot be nil[T ], since that would violate property 5 at p[x]. ● 8 cases in all, 4 of which are symmetric to the other.
    • Case 1 – w is red p[x] B w x A  D B D  x C  E     A  E new w   C   ● w must have black children. ● Make w black and p[x] red (because w is red p[x] couldn’t have been red). ● Then left rotate on p[x]. ● New sibling of x was a child of w before rotation  must be black. ● Go immediately to case 2, 3, or 4.
    • Case 2 – w is black, both w’s children p[x] are black new x c B w x A  B D A   C  E   c D  C   E   ● Take 1 black off x ( singly black) and off w ( red). ● Move that black to p[x]. ● Do the next iteration with p[x] as the new x. ● If entered this case from case 1, then p[x] was red  new x is red & black  color attribute of new x is RED  loop terminates. Then new x is made black in the last line. 
    • Case 3 – w is black, w’s left child is c red, w’s right child is black c B w x A  B x D A   C  E   new w  C  D   E  ● Make w red and w’s left child black. ● Then right rotate on w. ● New sibling w of x is black with a red right child  case 4. 
    • Case 4 – w is black, w’s right child is red c B w x A  D B D  C  c’   x E   E A  C     ● Make w be p[x]’s color (c). ● Make p[x] black and w’s right child black. ● Then left rotate on p[x]. ● Remove the extra black on x ( x is now singly black) without violating any red-black properties. ● All done. Setting x to root causes the loop to terminate.