Red-Black TreesPowerPoint PresentationSlide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Bottom –Up InsertionBottom Up InsertionSlide 13Slide 14Slide 15Slide 16Insertion PracticeAsymptotic Cost of Insertion1Red-Black Trees2A Red-Black Tree with NULLs shownBlack-Height of the tree = 43Red-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 NULL contains the same number of black nodes.–The root is black•Definition: The black-height of a node, X, in a red-black tree is the number of black nodes on any path to a NULL, not counting X.4A valid Red-Black TreeBlack-Height = 2567Theorem 1 – Any red-black tree with root x, has n >= 2bh(x) – 1 nodes, where bh(x) is the black height of node x.Proof: by induction on height of x.8Theorem 2 – In a red-black tree, at least half the nodes on any path from the root to a NULL must be black.Proof – If there is a red node on the path, there must be a corresponding black node.9Theorem 3 – In a red-black tree, no path from any node, N, to a NULL is more than twice as long as any other path from N to any other NULL.Proof: By definition, every path from a node to any NULL 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 NULL as on any other path. Therefore the length of every path is no more than twice as long as any other path10Theorem 4 –A red-black tree with n 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/2From Theorem 1, n >= 2bh(x) - 1Therefore n >= 2 h/2 – 1n + 1 >= 2h/2lg(n + 1) >= h/22lg(n + 1) >= h11Bottom –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–NULLs are Black–If node is Red, both children must be Black–Every path from node to descendant NULL must contain the same number of Blacks12Bottom Up Insertion•Insert node; Color it RED; X is pointer to it•Cases0: X is the root -- color it black1: Both parent and uncle are red -- color parent and uncle black, color grandparent red, point X to grandparent, check new situation2 (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(right) on parent, rotate right(left) on grandparent3 (zig-zig): Parent is red, but uncle is black. X and its parent are both left (right) children -- color parent black, color grandparent red, rotate right(left) on grandparent13XPGUPGUCase 1 – U is RedJust Recolor and move upX14XPGUSXPGSUCase 2 – Zig-ZagDouble Rotate X around P; X around GRecolor G and X15XPGUSPXGSUCase 3 – Zig-ZigSingle Rotate P around GRecolor P and G1611141521758Black nodeRed nodeInsert 4 into this R-B Tree17Insertion PracticeInsert the values 2, 1, 4, 5, 9, 3, 6, 7 into an initially empty Red-Black Tree18Asymptotic 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
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