3. 3
Red-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 leaf contains the same
number of black nodes.
• Definition: The black-height of a node, n, in
a red-black tree is the number of black nodes
on any path to a leaf, not counting n.
7. 7
Theorem 1 – Any red-black tree with root x,
has at least n = 2bh(x)
– 1 internal nodes,
where bh(x) is the black height of node x.
Proof: by induction on height of x.
8. 8
Theorem 2 – In a red-black tree, at least half
the nodes on any path from the root to a leaf
must be black.
Proof – If there is a red node on the path,
there must be a corresponding black node.
9. 9
Theorem 3 – In a red-black tree, no path from any
node, N, to a leaf is more than twice as long as any
other path from N to any other leaf.
Proof: By definition, every path from a node to any
leaf 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 leaf as on
any other path. Therefore the length of every path
is no more than twice as long as any other path
10. 10
Theorem 4 –
A red-black tree with n internal 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/2
From Theorem 1, n >= 2bh(x)
- 1
Therefore n >= 2 h/2
– 1
n + 1 >= 2h/2
lg(n + 1) >= h/2
2lg(n + 1) >= h
11. 11
Bottom –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
– Leaf nodes are Black NULLS
– If node is Red, both children must be Black
– Every path from node to descendant leaf must
contain the same number of Blacks
12. 12
Bottom Up Insertion
• Insert node; Color it RED; X is pointer to it
• Cases
0: X is the root -- color it black
1: Both parent and uncle are red -- color parent and uncle
black, color grandparent red, point X to grandparent,
check new situation
2 (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 on parent, rotate right on
grandparent
3 (zig-zig): Parent is red, but uncle is black. X and its
parent are both left or both right children -- color parent
black, color grandparent red, rotate right on
grandparent
18. 18
Asymptotic 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 n)
20. Recall “ordinary” BST Delete
• If vertex to be deleted is a leaf, just delete
it.
• If vertex to be deleted has just one child,
replace it with that child
• If vertex to be deleted has two children,
replace the value of by it’s in-order
predecessor’s value then delete the in-
order predecessor (a recursive step)
21. Bottom-Up Deletion
1. Do ordinary BST deletion. Eventually a
“case 1” or “case 2“ will be done (leaf or
just one child). If deleted node, U, is a
leaf, think of deletion as replacing with
the NULL pointer, V. If U had one child,
V, think of deletion as replacing U with V.
2. What can go wrong??
22. Which RB Property may be
violated after deletion?
1. If U is red?
Not a problem – no RB properties violated
2. If U is black?
If U is not the root, deleting it will change
the black-height along some path
23. Fixing the problem
• Think of V as having an “extra” unit of
blackness. This extra blackness must be
absorbed into the tree (by a red node), or
propagated up to the root and out of the
tree.
• There are four cases – our examples and
“rules” assume that V is a left child. There
are symmetric cases for V as a right child
24. Terminology
• The node just deleted was U
• The node that replaces it is V, which has an
extra unit of blackness
• The parent of V is P
• The sibling of V is S
Black Node
Red Node
Red or Black and don’t care
25. Bottom-Up Deletion
Case 1
• V’s sibling, S, is Red
– Rotate S around P and recolor S & P
• NOT a terminal case – One of the other
cases will now apply
• All other cases apply when S is Black
27. Bottom-Up Deletion
Case 2
• V’s sibling, S, is black and has two black
children.
– Recolor S to be Red
– P absorbs V’s extra blackness
• If P is Red, we’re done
• If P is Black, it now has extra blackness and
problem has been propagated up the tree
29. Bottom-Up Deletion
Case 3
• S is black
• S’s RIGHT child is RED (Left child either
color)
– Rotate S around P
– Swap colors of S and P, and color S’s Right
child Black
• This is the terminal case – we’re done
31. Bottom-Up Deletion
Case 4
• S is Black, S’s right child is Black and S’s
left child is Red
– Rotate S’s left child around S
– Swap color of S and S’s left child
– Now in case 3
35. Review of Bottom-Up Insertion
• In B-Up insertion, “ordinary” BST insertion
was used, followed by correction of the tree
on the way back up to the root
• This is most easily done recursively
– Insert winds up the recursion on the way down
the tree to the insertion point
– Fixing the tree occurs as the recursion unwinds
36. Top-Down Insertion Strategy
• In T-Down insertion, the corrections are
done while traversing down the tree to the
insertion point.
• When the actual insertion is done, no
further corrections are needed, so no need
to traverse back up the tree.
• So, T-Down insertion can be done
iteratively which is generally faster
37. Goal of T-D Insertion
• Insertion is always done as a leaf (as in
ordinary BST insertion)
• Recall from the B-Up flow chart that if the
uncle of a newly inserted node is black, we
restore the RB tree properties by one or two
local rotations and recoloring – we do not
need to make changes further up the tree
38. Goal (2)
• Therefore, the goal of T-D insertion is to
traverse from the root to the insertion point
in such a way that RB properties are
maintained, and at the insertion point, the
uncle is Black.
• That way we may have to rotate and
recolor, but not propagate back up the tree
39. Possible insertion configurations
X (Red or Black)
Y Z
If a new node is inserted as a child of Y or Z, there is no
problem since the new node’s parent is black
40. Possible insertion configurations
X
Y Z
If new node is child of Z, no problem since Z is black.
If new node is child of Y, no problem since the new node’s
uncle (Z) is black – do a few rotations and recolor…. done
41. Possible insertion configurations
X
Y Z
If new node is inserted as child of Y or Z, it’s uncle will
be red and we will have to go back up the tree. This is the
only case we need to avoid.
42. Top-Down Traversal
X
Y Z
As we traverse down the tree and
encounter this case, we recolor and
possible do some rotations.
There are 3 cases.
Remember the goal – to create an insertion point at which the
parent of the new node is Black, or the uncle of the new node
is black.
43. Case 1 – X’s Parent is Black
X
Z
Y
P
X
Z
P
Just recolor and continue down the tree
Y
44. Case 2
• X’s Parent is Red (so Grandparent is Black)
and X and P are both left/right children
– Rotate P around G
– Color P black
– Color G red
• Note that X’s uncle, U, must be black
because it (a) was initially black, or (b)
would have been made black when we
encountered G (which would have had two
red children -- X’s Parent and X’s uncle)
46. Case 3
• X’s Parent is Red (so Grandparent is Black)
and X and P are opposite children
– Rotate P around G
– Color P black
– Color G red
• Again note that X’s uncle, U, must be black
because it (a) was initially black, or (b)
would have been made black when we
encountered G (which would have had two
red children -- X’s Parent and X’s uncle)
47. Case 3 Diagrams (1 of 2)
X
Z
Y
P
G
U
S
X
Y
S
P
G
U
Z
Step 1 – recolor X, Y and Z. Rotate X around P.
48. Case 3 Diagrams (2 of 2)
X
Y
S
P
G
U
Z
P
Y
S
X
G
U
Z
Step 2 – Rotate X around G. Recolor X and G
50. Top-Down Insert Summary
P
X
Y Z
Case 1
P is Black
Just Recolor
P
X
Y Z
Case 2
P is Red
X & P both left/right
P
X
Y Z
G
P
X
Y Z
G
Recolor
X,Y,Z
P
X
Y Z
G
Rotate P
around G
Recolor P,G
Case 3
P is Red
X and P are
opposite children
P
X
Y Z
G
Recolor X,Y,Z
Rotate X
around P
X
P
Y Z
G
Rotate X
around G
Recolor X, G
Recolor
X,Y,Z
X
P
Y Z
G
52. Recall the rules for BST deletion
• If vertex to be deleted is a leaf, just delete
it.
• If vertex to be deleted has just one child,
replace it with that child
• If vertex to be deleted has two children,
replace the value of by it’s in-order
predecessor’s value then delete the in-
order predecessor (a recursive step)
53. What can go wrong?
1. If the delete node is red?
Not a problem – no RB properties violated
2. If the deleted node is black?
If the node is not the root, deleting it will
change the black-height along some path
54. The goal of T-D Deletion
• To delete a red leaf
• How do we ensure that’s what happens?
– As we traverse the tree looking for the leaf to
delete, we change every node we encounter to
red.
– If this causes a violation of the RB properties,
we fix it
55. Bottom-Up vs. Top-Down
• Bottom-Up is recursive
– BST deletion going down the tree (winding up
the recursion)
– Fixing the RB properties coming back up the
tree (unwinding the recursion)
• Top-Down is iterative
– Restructure the tree on the way down so we
don’t have to go back up
56. Terminology
• Matching Weiss text section 12.2
– X is the node being examined
– T is X’s sibling
– P is X’s (and T’s) parent
– R is T’s right child
– L is T’s left child
• This discussion assumes X is the left child of P.
As usual, there are left-right symmetric cases.
57. Basic Strategy
• As we traverse the tree, we change every
node we visit, X, to Red.
• When we change X to Red, we know
– P is also Red (we just came from there)
– T is black (since P is Red, it’s children are
Black)
58. Step 1 – Examine the root
1. If both of the root’s children are Black
a. Make the root Red
b. Move X to the appropriate child of the root
c. Proceed to step 2
2. Otherwise designate the root as X and
proceed to step 2B.
59. Step 2 – the main case
As we traverse down the tree, we continually
encounter this situation until we reach the
node to be deleted
X is Black, P is Red, T is Black
We are going to color X Red, then recolor
other nodes and possibly do rotation(s)
based on the color of X’s and T’s children
2A. X has 2 Black children
2B. X has at least one Red child
60. P
T
X
Case 2A
X has two Black Children
2A1. T has 2 Black Children
2A2. T’s left child is Red
2A3. T’s right child is Red
** if both of T’s children are Red,
we can do either 2A2 or 2A3
61. Case 2A1
X and T have 2 Black Children
P
T
X
P
T
X
Just recolor X, P and T and move down the tree
62. Case 2A2
P
T
X
L
X has 2 Black Children and T’s Left Child is Red
Rotate L around T, then L around P
Recolor X and P then continue down the tree
L1 L2
P T
X
L
L1 L2
63. Case 2A3
P
T
X
X has 2 Black Children and T’s Right Child is Red
Rotate T around P
Recolor X, P, T and R then continue down the tree
R1 R2
P R
X
T
R2
R1
R
L L
64. Case 2B
X has at least one Red child
Continue down the tree to the next level
If the new X is Red, continue down again
If the new X is Black (T is Red, P is Black)
Rotate T around P
Recolor P and T
Back to main case – step 2
65. Case 2B Diagram
P
X T
Move down the tree.
P
X T
P
T X
If move to Black child (2B2)
Rotate T around P; Recolor P and T
Back to step 2, the main case
If move to the Red child (2B1)
Move down again
66. Step 3
Eventually, find the node to be deleted – a leaf or a node with
one non-null child that is a leaf.
Delete the appropriate node as a Red leaf
Step 4
Color the Root Black
67. Example 1
Delete 10 from this RB Tree
15
17
16
20
23
18
13
10
7
12
6
3
Step 1 – Root has 2 Black children. Color Root Red
Descend the tree, moving X to 6
71. Example 2
Delete 10 from this RB Tree
15
17
16
20
13
10
12
6
3
4
2
Step 1 – the root does not have 2 Black children.
Color the root red, Set X = root and proceed to step 2
74. Example 2 (cont’d)
15
17
16
20
13
10
12
6
3
4
2
P
X T
X is now the leaf to be deleted, but it’s Black, so back to step 2.
X has 2 Black children and T has 2 Black children – case 2A1
Recolor X, P and T.
Step 3 -- Now delete 10 as a red leaf.
Step 4 -- Recolor the root black
76. Example 3
Delete 11 from this RB Tree
15
13
11
12
10
5
7
3
6 9
2 4
Valid and
unaffected
Right subtree
Step 1 – root has 2 Black children. Color Root red.
Set X to appropriate child of root (10)
78. Example 3 (cont’d)
15
13
11
12
10
5
7
3
6 9
4
X
Since we arrived at a black node (case 2B2) assuring T is red
and P is black), rotate T around P, recolor T and P
Back to step 2
P
T
2
79. Example 3 (cont’d)
15
13
11
12
10
5
7
3
6 9
4
X
P
T
2
Now X is Black with Red parent and Black sibling.
X and T both have 2 Black children (case 2A1)
Just recolor X, P and T and continue traversal
80. Example 3 (cont’d)
15
13
11
12
10
5
7
3
6 9
4
X
P
T
2
Having traversed down the tree, we arrive at 11, the leaf to
be deleted, but it’s Black, so back to step 2.
X and T both have two Black children. Recolor X, P and T.
Step 3 -- delete 11 as a red leaf.
Step 4 -- Recolor the root black
