DeleteDelete 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.
Bottom-Up Red-Black Trees
Top-down red-black trees require O(log n) rotations per insert/delete.
Color flips cheaper than rotations.
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.
Bottom-Up Red-Black Trees
Bottom-up red-black tree properties.
At most 1 rotation per insert/delete.
O(1) amortized complexity to restructure following an insert/delete.
Bottom-Up Insert
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
a b c d gp pp p XYz
X => relationship between gp and pp.
pp left child of gp => X = L.
Y => relationship between pp and p.
p right child of pp => Y = R.
z = b (black) if d = null or a black node.
z = r (red) if d is a red node.
If p is a red root, make it black.
If p or pp black done!
If pp is a red root, make it black.
O.w., gp exists
XYr
a b c d gp pp p a b c d gp pp p Move p, pp, and gp up two levels.
Continue rebalancing if necessary. Color flip.
LLb
Done!
Same as LL rotation of AVL tree. y x a b z c d a b c d gp pp p x y z Rotate.
LRb
Done!
Same as LR rotation of AVL tree.
RRb and RLb are symmetric. y x a b z c d b c a d gp pp p y x z Rotate.
Exercise: build the red-black tree
Build a red-black tree from the following key values in sequence: 45, 28, 36, 22, 68, 50, 10, 98, 26, 15, 25, 27, 78
Delete
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 Leaf
10 7 8 1 5 30 40 20 25 35 45 60 3 Delete 8.
Delete A Black Leaf
y y is root of deficient subtree.
py is parent of y. 10 7 1 5 30 40 20 25 35 45 60 3 py
If py doesn’t exist, the whole tree is one black node deficient. This is OK.
Delete A Black Degree 1 Node
10 7 8 1 5 30 40 20 25 35 45 60 3 Delete 45. y y is root of deficient subtree. py
Delete A Black Degree 2 Node
10 7 8 1 5 30 40 20 25 35 45 60 3 Not possible, degree 2 nodes are never deleted.
Rebalancing Strategy
10 7 8 1 5 30 40 20 25 35 45 60 3 y py If y is a red node, make it black.
Rebalancing Strategy
60 10 7 8 1 5 30 40 20 25 35 45 3 y py Now, no subtree is deficient. Done!
Rebalancing Strategy
60 10 7 8 1 5 30 40 20 25 35 45 3 y y is a black root (there is no py).
Entire tree is deficient. Done!
Rebalancing Strategy
Xcn
y is right child of py => X = R.
Pointer to v is black => c = b.
v has 1 red child => n = 1. a b y py v y is black but not the root (there is a py).
y represents both a subtree and its root.
Broken lines indicate pointers whose colors are unspecified.
v must exist because y is black.
Rb0 (case 1)
a b y py v y a b py v Color change.
Now, py is root of deficient subtree.
Continue!
Rb0 (case 2)
a b y py v y a b py v Color change.
Deficiency eliminated.
Done!
Rb1 (case 1)
a b y py v a b y v py LL rotation.
Deficiency eliminated.
Done!
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