Red Black TreesSlide 2Example Red-Black TreePropertiesSlide 5Slide 6Slide 7Slide 8Slide 9InsertClassification Of 2 Red Nodes/PointersXYrLLbLRbDeleteDelete A Black LeafSlide 17Delete A Black Degree 1 NodeDelete A Black Degree 2 NodeRebalancing StrategySlide 21Slide 22Slide 23Rb0 (case 1, py is black)Rb0 (case 2, py is red)Rb1 (case 1)Rb1 (case 2)Rb2Rr(n)Rr(0)Rr(1) (case 1)Rr(1) (case 2)Rr(2)Red Black TreesColored Nodes Definition•Binary search tree.•Each node is colored red or black.•Root and all external nodes are black.•No root-to-external-node path has two consecutive red nodes.•All root-to-external-node paths have the same number of black nodesRed Black TreesColored Edges Definition•Binary search tree.•Child pointers are colored red or black.•Pointer to an external node is black.•No root to external node path has two consecutive red pointers.•Every root to external node path has the same number of black pointers.Example Red-Black Tree107815304020253545603Properties•The height of a red black tree that has n (internal) nodes is between log2(n+1) and 2log2(n+1).Properties•Start with a red black tree whose height is h; collapse all red nodes into their parent black nodes to get a tree whose node-degrees are between 2 and 4, height is >= h/2, and all external nodes are at the same level.107815304020253545603PropertiesProperties•Let h’>= h/2 be the height of the collapsed tree. •Internal nodes of collapsed tree have degree between 2 and 4.•Number of internal nodes in collapsed tree >= 2h’-1.•So, n >= 2h’-1•So, h <= 2 log2 (n + 1)Properties•At most 1 rotation and O(log n) color flips per insert/delete.•Priority search trees.Two keys per element.Search tree on one key, priority queue on other.Color flip doesn’t disturb priority queue property.Rotation disturbs priority queue property.O(log n) fix time per rotation => O(log2n) overall time for AVL.Properties•O(1) amortized complexity to restructure following an insert/delete.•C++ STL implementation•java.util.TreeMap => red black treeInsert•New pair is placed in a new node, which is inserted into the red-black tree.•New node color options.Black node => one root-to-external-node path has an extra black node (black pointer).•Hard to remedy.Red node => one root-to-external-node path may have two consecutive red nodes (pointers).•May be remedied by color flips and/or a rotation.Classification Of 2 Red Nodes/Pointers•XYzX => relationship between gp and pp.•pp left child of gp => X = L.Y => relationship between pp and p.•p left child of pp => Y = L.z = b (black) if d = null or a black node.z = r (red) if d is a red node.a bcdgppppXYr•Color flip.a bcdgppppa bcdgpppp•Move p, pp, and gp up two levels.•Continue rebalancing.LLb•Rotate.•Done!•Same as LL rotation of AVL tree.yxa bzc da bcdgppppxyzLRb•Rotate.•Done!•Same as LR rotation of AVL tree.•RRb and RLb are symmetric.yxa bzc db cadgppppyxzDelete•Delete as for unbalanced binary search tree.•If red node deleted, no rebalancing needed.•If black node deleted, a subtree becomes one black pointer (node) deficient.Delete A Black Leaf107815304020253545603• Delete 8.Delete A Black Leafy• y is root of deficient subtree.• py is parent of y.10715304020253545603pyDelete A Black Degree 1 Node107815304020253545603• Delete 45.y• y is root of deficient subtree.pyDelete A Black Degree 2 Node107815304020253545603• Not possible, degree 2 nodes are never deleted.Rebalancing Strategy•If y is a red node, make it black.107815304020253545603ypyRebalancing Strategy•Now, no subtree is deficient. Done!601078153040202535453ypyRebalancing Strategy•y is a black root (there is no py).•Entire tree is deficient. Done!601078153040202535453yRebalancing Strategy•y is black but not the root (there is a py).•Xcny is right child of py => X = R.Pointer to v is black => c = b.v has 1 red child => n = 1.a bypyvRb0 (case 1, py is black)•Color change.•Now, py is root of deficient subtree.•Continue!a bypyvya bpyvRb0 (case 2, py is red)•Color change.•Deficiency eliminated.•Done!a bypyvya bpyvRb1 (case 1)•LL rotation.•Deficiency eliminated.•Done!a bypyvab yvpyRb1 (case 2)•LR rotation.•Deficiency eliminated.•Done!aypyvb cwc ywpya bvRb2•LR rotation.•Deficiency eliminated.•Done!aypyvb cwc ywpya bvRr(n)•n = # of red children of v’s right child w.aypyvb cwRr(0)•LL rotation.•Done!a bypyvab yvpyRr(1) (case 1)•LR rotation.•Deficiency eliminated.•Done!aypyvbwcycwpyavbRr(1) (case 2)•Rotation.•Deficiency eliminated.•Done!aypyvbwc dxydxpyavb cwRr(2)•Rotation.•Deficiency eliminated.•Done!aypyvbwc dxd yxpyavb