- Properties of a
Red-Black tree:
- Each node in the
binary search tree is
labeled either as
red or as
black
- The root node is
always
labeled
as black
- All the leaf nodes are
always
labeled
as black
- Every red node must have
2 black children nodes
(There is no child constraint on black (parent) nodes)
- Every path from root node → leaf node must have the same number of black nodes
- Each node in the
binary search tree is
labeled either as
red or as
black
- When the node N is
inserted
into a Red-Black tree,
we will (always) end up with
one of the following
6 cases:
(Recall that the color of the new node N is always red)
-
Node N is the
root node
(i.e.:
node N has
no parent node)
- The parent node P of
new node N is
black:
- The parent node P of
new node N is
red
and the
sibling of P is
red:
(There are 2 symmetric cases in this case)
- The parent node P of
new node N is
red
and the
sibling of P is
black:
(There are 2 symmetric cases in this case)
We will next discuss how to adjust the resulting tree to restore all the properties of the Red-Black tree for each of the above cases
(Please convince yourself that these are all the possible cases)
-
Node N is the
root node
(i.e.:
node N has
no parent node)
- Case 1:
newly inserted node N is the
root node
(i.e.:
node N has
no parent node)
- Violation:
- Property #2: The root node is always labeled as black
- Fix:
- Change the
color of
node N to
black:
And we are done !!!
- Change the
color of
node N to
black:
- Case 2:
the parent node P of
new node N is
black:
- Recall that a
new node is
always inserted in a
leaf node position and
the leaf nodes are
black:
Notice also that:
- We inserted a
(new) red node N, so
the
number of black nodes in
a path is
unchanged !!!
- The (new) red node N has a black parent node and a black child node (i.e.: we did not create a situation with 2 consecutive red nodes)
- We inserted a
(new) red node N, so
the
number of black nodes in
a path is
unchanged !!!
- Therefore:
- The new tree will satisfy all properties of the Red-Black tree !!!
So we are done !!!
- Case 3:
the parent node P of
new node N is
red
and the
sibling of P is
red:
- Violation:
- Property #4:
Every red node must have
2 black children nodes
(Property #5 is preserved because we did not introduce any black nodes !!!)
- Property #4:
Every red node must have
2 black children nodes
- Situation:
(I will use
case (1) to
illustrate the
re-adjustment technique -
case 2 is
symmetrical)
- State
after inserting the
new (red) node N:
- Because the original Red-Black tree does
not have
2 consecutive red nodes,
the sub-trees of
P and U must have
black root nodes:
- State
after inserting the
new (red) node N:
- Fix:
- We re-color the
nodes
G,
P and
U as follows:
Notice that:
- The number of black nodes on
any path is
unchanged !!!
- We have removed the violation: node P is now black and we don't have 2 consecutive red nodes in any path
- The number of black nodes on
any path is
unchanged !!!
However:
- The red node G
can cause (a new)
property #4
violation
if
its parent node is
red:
- Fortunately, this situation is exactly the same case as when we inserted the new node N:
- Fortunately, this situation is exactly the same case as when we inserted the new node N:
Therefore, we can use recursion (i.e.: re-coloring) on the node G to fix the potential violation at node G in the Red-Black tree !!!
- We re-color the
nodes
G,
P and
U as follows:
- Case 4:
the parent node P of
new node N is
red
and the
sibling of P is
black:
- Violation:
- Property #4:
Every red node must have
2 black children nodes
(Property #5 is preserved because we did not introduce any black nodes !!!)
- Property #4:
Every red node must have
2 black children nodes
- Situation (A):
(I will use
case (1a) to
illustrate the
re-adjustment technique -
case 2a is
symmetrical)
- State
after inserting the
new (red) node N:
- Because the original Red-Black tree does
not have
2 consecutive red nodes,
the sub-trees of
P and U must have
black root nodes:
- State
after inserting the
new (red) node N:
- Fix:
- We first perform the
following
right tree-rotation operation
that preserves the
properties of
a Binary Search tree:
- The resulting tree after
applying the
right tree-rotation operation is
as follows:
Notice that:
- The number of black nodes on
path P → N → leaf is
reduced by 1 !!!
- The resulting tree still has
2 consecutive red nodes !!!
- Both violations can be fixed with a re-coloring operation !!!
- The number of black nodes on
path P → N → leaf is
reduced by 1 !!!
- Step 2:
re-color the
nodes P and G:
Notice that:
- The number of black nodes on
every path is
unchanged
(equal to the original tree) !!!
- The resulting tree has no consecutive red nodes on any path !!!
We are done !
- The number of black nodes on
every path is
unchanged
(equal to the original tree) !!!
- We first perform the
following
right tree-rotation operation
that preserves the
properties of
a Binary Search tree:
- Situation (B):
(I will use
case (1b) to
illustrate the
re-adjustment technique -
case 2b is
symmetrical)
- State
after inserting the
new (red) node N:
- Because the original Red-Black tree does
not have
2 consecutive red nodes,
the sub-trees of
P and U must have
black root nodes:
- State
after inserting the
new (red) node N:
- Fix:
- We perform the
following
left tree-rotation operation
that preserves the
properties of
a Binary Search tree
on the node P:
- The resulting tree after
applying the
left tree-rotation operation is
as follows:
The new tree is a case 4 - (1a) --- which we have solved already !!!
- We perform the
following
left tree-rotation operation
that preserves the
properties of
a Binary Search tree
on the node P: