Introduction to Algorithms
Red-Black Trees
My T. Thai @ UF
Red-Black Trees
 A binary search tree, each node is colored red
or back
 Red-black properties:
1. Every node is either red or black
2. The root is black
3. Every leaf (NIL) is black
4. If a node is red, then both its children are black
5. All simple paths from any node x to descendant
leaves contain the same number of black nodes
= black-height (bh(x))
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Example
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Key property
Proof:
 Claim 1: The subtree rooted at x (internal(x)) contains
at least 2bh(x)-1 internal nodes
 Proof: Use induction and the property: internal(x) =
internal(x.left) + internal(x .right)
 Claim 2: Any node of height h has black-height
 Proof: use property 4
 From Claim 1&2: h(root)
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2bh(root)
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Operator
 Querying operators: Search, Minimum,
Maximum, Successor, Predecessor are the
same as binary search tree
 Time: O(log n)
 Updating operators: need to modify to
guarantee red-black properties:
 Insertion, Deletion: may violates some properties
 Rotations: tree-reconstructing operation that
maintain the properties of red-black tree
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Rotations




left and right rotations: inverse of each other
Keep the properties of binary search tree
Change the tree structure
Time: O(1)
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Example
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Left rotation
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Insertion
 Key problem inserting z:
 What is the color of inserted node?
 The properties of red-black tree may be violated
 Solution:
 Insert z using the same procedure on binary search
tree
 Fix violated properties:
 Change the structure of the tree using rotations
 Recolor nodes
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Insertion
 Possible violations of
properties
1.OK
2.If z is root: easy to fix
3.OK
4.If z.p is red: both z and z.p are
red
5.OK
 Change tree and fix property 4
iteratively
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Fix tree
 Case 1: z’s uncle y is red
 Change color of z.p, z.p.p and y then fix z.p.p
z.p.p must
be black
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Fix tree
 Case 2: y is black, z is the right child
 Left rotate at z.p  case 3
 Case 3: y is black, z is the left child
 Make z.p black, z.p.p red then rotate on z.p.p
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Fix tree
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Running time of Insertion
 Inserting new node using insert procedure (on
binary search tree) takes O(log n)
 Fixing time: O(log n)
 Fix case 1 iteratively takes O(log n) time
 Fix case 2, 3: constant time
 Total running time: O(log n)
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Deletion
 Delete node z using the same
procedure on binary search tree
 y is the removed node
 Possible violations if y is
black:
2.y is root and x is red  red root
4.y.p and x are both red
5.Any path containing y has 1 fewer
black node
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Fix property 5
 Correct by giving x an “extra black”
 Violate property 1: x points to either doubly
black (if x.color = BLACK) or red & black (if
x.color = RED) node
 Solution:




x.color = RED  turn it into a black node
x points to the root  remove the extra black
x always points to a non-root doubly black node
w is x’s sibling, w is not a T.nil  4 cases
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Fix tree
 Case 1: w is red
 w has black children  sibling of x is black 
cases 2, 3, 4
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 Case 2: w is black and both of w’s children are
black
 Take 1 black off x (⇒ singly black) and off w
(⇒red)
 Move black to x.p
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 Case 3: w is black, w’s left child is red, and w’s
right child is black
 Exchange color of w and w’s left child then rotate
 Sibling of x is black  case 4
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 Case 4: w is black, w’s left child is black, and
w’s right child is red
 Make w be x.p’s color then make x.p and w’s
right black
 Rotate on x.p then remove extra black on x
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RB-DELETE-FIXUP
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Running time of Deletion
 Delete a node using delete procedure (on binary
search tree) takes O(log n)
 Fixing time: O(log n)
 Fix case 2 iteratively takes O(log n) time (in each
iteration x moves up 1 level)
 Fix case 1, 3, 4: constant time
 Total running time: O(log n)
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