Initially, a new item is inserted just as in a binary search tree. Note that the item always goes into a new leaf. The tree is then readjusted as needed in order to maintain it as an AVL tree. There are three main cases to consider when inserting a new node.
Algo : Single Right Rotation OR RR rotation
(L.w. - shedding load from left to RIGHT) ( p = k2)
temp = p->left;
p->left = temp->right;
temp->right = p;
p = temp;
Consider the following more detailed example that illustrates subcase A.
So tree acquires this shape:
The fix is to use a double right rotation at node P. A double right rotation at P consists of a single left rotation at LC followed by a single right rotation at P. (In the mirror image case a double left rotation is used at P. This consists of a single right rotation at the right child RC followed by a single left rotation at P.) In the above picture, the double rotation gives the following (where we first show the result of the left rotation at LC, then a new picture for the result of the right rotation at P).
Case 1: New Node added to (BF = 0) Node
A node with balance factor 0 changes to +1 or -1 when a new node is inserted below it. No change is needed at this node. Consider the following example. Note that after an insertion one only needs to check the balances along the path from the new leaf to the root.
0
40
/ \
+1 0
20 50
\ / \
0 0 0
30 45 70
After inserting 60 we get:
+1
40
/ \
+1 +1
20 50
\ / \
0 0 -1
30 45 70
/
0
60
Case 2: Node added to right of BF=-1 node or left of BF=+1 Node
A node with balance factor -1 changes to 0 when a new node is inserted in its right subtree. (Similarly for +1 changing to 0 when inserting in the left subtree.) No change is needed at this node. Consider the following example.
-1
40
/ \
+1 0
20 50
/ \ / \
0 0 0 0
10 30 45 70
/ \
0 0
22 32
After inserting 60 we get: 0 <-- the -1 changed to a 0 (case 2)
40
/ \
+1 +1 <-- an example of case 1
20 50
/ \ / \
0 0 0 -1 <-- an example of case 1
10 30 45 70
/ \ /
0 0 0
22 32 60
Case 3:
A node with balance factor -1 changes to -2 when a new node is inserted in its left subtree. (Similarly for +1 changing to +2 when inserting in the right subtree.) Change is needed at this node. The tree is restored to an AVL tree by using a rotation.Subcase A:
This consists of the following situation, where P denotes the parent of the subtree being examined, LC is P's left child, and X is the new node added. Note that inserting X makes P have a balance factor of -2 and LC have a balance factor of -1. The -2 must be fixed. This is accomplished by doing a right rotation at P. Note that rotations do not mess up the order of the nodes given in an inorder traversal. This is very important since it means that we still have a legitimate binary search tree. (Note, too, that the mirror image situation is also included under subcase A.)
(rest of tree)
|
-2
P
/ \
-1 sub
LC tree
of
/ \ height
n
sub sub
tree tree
of of
height height
n n
/
X
The fix is to use a single right rotation at node P. (In the mirror image case a single left rotation is used at P.) This gives the following picture.
(rest of tree)
|
0
LC
/ \
sub P
tree
of / \
height
n sub sub
/ tree tree
X of of
height height
n n
Putting it diagramatically:
(L.w. - shedding load from left to RIGHT) ( p = k2)
temp = p->left;
p->left = temp->right;
temp->right = p;
p = temp;
Consider the following more detailed example that illustrates subcase A.
-1
80
/ \
-1 -1
30 100
/ \ /
0 0 0
15 40 90
/ \
0 0
10 20
We then insert 5 and then check the balance factors from the new leaf up toward the root. (Always check from the bottom up.)
-2
80
/ \
-2 -1
30 100
/ \ /
-1 0 0
15 40 90
/ \
-1 0
10 20
/
0
5
This reveals a balance factor of -2 at node 30 that must be fixed. (Since we work bottom up, we reach the -2 at 30 first. The other -2 problem will go away once we fix the problem at 30.) The fix is accomplished with a right rotation at node 30, leading to the following picture.
-1
80
/ \
0 -1
15 100
/ \ /
-1 0 0
10 30 90
/ / \
0 0 0
5 20 40
Recall that the mirror image situation is also included under subcase A. The following is a general illustration of this situation. The fix is to use a single left rotation at P. See if you can draw a picture of the following after the left rotation at P. Then draw a picture of a particular example that fits our general picture below and fix it with a left rotation.
(rest of tree)
|
+2
P
/ \
sub +1
tree RC
of
height / \
n
sub sub
tree tree
of of
height height
n n
\
X
Subcase B:
This consists of the following situation, where P denotes the parent of the subtree being examined, LC is P's left child, NP is the node that will be the new parent, and X is the new node added. X might be added to either of the subtrees of height n-1. Note that inserting X makes P have a balance factor of -2 and LC have a balance factor of +1. The -2 must be fixed. This is accomplished by doing a double rotation at P (explained below). (Note that the mirror image situation is also included under subcase B.)
(rest of tree)
|
-2
P
/ \
+1 sub
LC tree
of
/ \ height
n
sub -1
tree NP
of / \
height sub sub
n tree tree
n-1 n-1
/
X
So tree acquires this shape:
The fix is to use a double right rotation at node P. A double right rotation at P consists of a single left rotation at LC followed by a single right rotation at P. (In the mirror image case a double left rotation is used at P. This consists of a single right rotation at the right child RC followed by a single left rotation at P.) In the above picture, the double rotation gives the following (where we first show the result of the left rotation at LC, then a new picture for the result of the right rotation at P).
(rest of tree)
|
-2
P
/ \
-2 sub
NP tree
of
/ \ height
n
0 sub
LC tree
/ \ n-1
sub sub
tree tree
of n-1
height /
n X
Finally we have the following picture after doing the right rotation at P. (rest of tree)
|
0
NP
/ \
0 +1
LC P
/ \ / \
sub sub sub sub
tree tree tree tree
of n-1 n-1 of
height / height
n X n
Consider the following concrete example of subcase B.
-1
80
/ \
0 0
30 100
/ \ / \
-1 0 0 0
20 50 90 120
/ / \
0 0 0
10 40 60
After inserting 55, we get a problem, a balance factor of -2 at the root node, as seen below.
-2
80
/ \
+1 0
30 100
/ \ / \
-1 +1 0 0
20 50 90 120
/ / \
0 0 -1
10 40 60
/
0
55
As discussed above, this calls for a double rotation. First we do a single left rotation at 30. This gives the following picture.
-2
80
/ \
-1 0
50 100
/ \ / \
-1 -1 0 0
30 60 90 120
/ \ /
-1 0 0
20 40 55
/
0
10
Finally, the right rotation at 80 restores the binary search tree to be an AVL tree. The resulting picture is shown below. 0
50
/ \
-1 0
30 80
/ \ / \
-1 0 -1 0
20 40 60 100
/ / / \
0 0 0 0
10 55 90 120
0 comments:
Post a Comment