Red Black TreeA Red Black Tree is a category of the selfbalancing binary search tree. It was created in 1972 by Rudolf Bayer who termed them "symmetric binary Btrees." A redblack tree is a Binary tree where a particular node has color as an extra attribute, either red or black. By check the node colors on any simple path from the root to a leaf, redblack trees secure that no such path is higher than twice as long as any other so that the tree is generally balanced. Properties of RedBlack TreesA redblack tree must satisfy these properties:
A tree T is an almost redblack tree (ARB tree) if the root is red, but other conditions above hold. Operations on RB Trees:The searchtree operations TREEINSERT and TREEDELETE, when runs on a redblack tree with n keys, take O (log n) time. Because they customize the tree, the conclusion may violate the redblack properties. To restore these properties, we must change the color of some of the nodes in the tree and also change the pointer structure. 1. Rotation:Restructuring operations on redblack trees can generally be expressed more clearly in details of the rotation operation. Clearly, the order (Ax By C) is preserved by the rotation operation. Therefore, if we start with a BST and only restructure using rotation, then we will still have a BST i.e. rotation do not break the BSTProperty. LEFT ROTATE (T, x) 1. y ← right [x] 1. y ← right [x] 2. right [x] ← left [y] 3. p [left[y]] ← x 4. p[y] ← p[x] 5. If p[x] = nil [T] then root [T] ← y else if x = left [p[x]] then left [p[x]] ← y else right [p[x]] ← y 6. left [y] ← x. 7. p [x] ← y. Example: Draw the complete binary tree of height 3 on the keys {1, 2, 3... 15}. Add the NIL leaves and color the nodes in three different ways such that the black heights of the resulting trees are: 2, 3 and 4. Solution: Tree with blackheight2 Tree with blackheight3 Tree with blackheight4 2. Insertion:
A discrepancy can decision from a parent and a child both having a red color. This type of discrepancy is determined by the location of the node concerning grandparent, and the color of the sibling of the parent. RBINSERT (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 14. left [z] ← nil [T] 15. right [z] ← nil [T] 16. color [z] ← RED 17. RBINSERTFIXUP (T, z) After the insert new node, Coloring this new node into black may violate the blackheight conditions and coloring this new node into red may violate coloring conditions i.e. root is black and red node has no red children. We know the blackheight violations are hard. So we color the node red. After this, if there is any color violation, then we have to correct them by an RBINSERTFIXUP procedure. RBINSERTFIXUP (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[z]] ← RED //Case 1 8. z ← p[p[z]] //Case 1 9. else if z= right [p[z]] 10. then z ← p [z] //Case 2 11. LEFTROTATE (T, z) //Case 2 12. color [p[z]] ← BLACK //Case 3 13. color [p [p[z]]] ← RED //Case 3 14. RIGHTROTATE (T,p [p[z]]) //Case 3 15. else (same as then clause) With "right" and "left" exchanged 16. color [root[T]] ← BLACK Example: Show the redblack trees that result after successively inserting the keys 41,38,31,12,19,8 into an initially empty redblack tree. Solution:Insert 41 Insert 19 Thus the final tree is 3. Deletion:First, search for an element to be deleted
The strategy RBDELETE is a minor change of the TREEDELETE procedure. After splicing out a node, it calls an auxiliary procedure RBDELETEFIXUP that changes colors and performs rotation to restore the redblack properties. RBDELETE (T, z) 1. if left [z] = nil [T] or right [z] = nil [T] 2. then y ← z 3. else y ← TREESUCCESSOR (z) 4. if left [y] ≠ nil [T] 5. then x ← left [y] 6. else x ← right [y] 7. p [x] ← p [y] 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 RBdeleteFIXUP (T, x) 18. return y RBDELETEFIXUP (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. LEFTROTATE (T, p [x]) //Case 1 8. w ← right [p[x]] //Case 1 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. RIGHTROTATE (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. LEFTROTATE (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 Example: In a previous example, we found that the redblack tree that results from successively inserting the keys 41,38,31,12,19,8 into an initially empty tree. Now show the redblack trees that result from the successful deletion of the keys in the order 8, 12, 19,31,38,41. Solution: Delete 38 Delete 41 No Tree.
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