Recall the definition of the AVL invariant
Definition The AVL Invariant: the height of two child subtrees may only differ by 1, for every node in the tree.
Do the normal BST insert.
Fix the AVL property if needed.
We may need to fix problems along the entire path from the point of insertion on up to the root.
Example insertion and rebalancing:
41
/ \
20 \
/ \ 65
11 29 /
/ 50
26
insert(23) ==>
41
/ \
20 \
/ \ 65
11 29 /
/ 50
26
/
23
Node 29 breaks the AVL invariant.
y x
/ \ right_rotate(y) / \
x C ---------------> A y
/ \ <------------- / \
A B left_rotate(x) B C
Rotations preserve the BST property and the in-order ordering.
A x B y C = A x B y C
Insert example: let’s use rotation to fix up our insert(23) example:
29
/ right_rotate(29)
26 ----------------> 26
/ / \
23 23 29
However, in different situations, the way in which we use tree rotation is different. So let’s look at more situations.
41
/ \
20 65
/ \ /
11 29 50
/
26
So 65 breaks the AVL invariant, and we have a zig-zag:
65
/
50
\
55
A right rotation at 65 gives us a zag-zig, we’re not making progress!
65(y) 50(x)
/ right_rotate(65) \
50(x) ----------------> 65(y)
\ /
55(B) 55(B)
Instead, let’s try a left rotate at 50:
65 65
/ left_rotate(50) /
50(x) ---------------> 55(y)
\ /
55(y) 50(x)
This looks familiar, now we can rotate right.
65(y) 55(x)
/ right_rotate(65) / \
55(x) ---------------> 50(A) 65(y)
/
50(A)
_6[2]_
/ \
3[0] 8[1]
/ \
7[0] 10[0]
Insert 11:
_6[3]_
/ \
3[0] 8[2]
/ \
7[0] 10[1]
\
11[0]
starting with an empty AVL tree, insert
14, 17, 11, 7, 4, 53, 13, 12, 8
Solution:
after insert 14, 17, 11, 7:
14
/ \
11 17
/
7
insert 4:
14
/ \
11 17
/
7
/
4
Node 11 doesn’t satisfy AVL.
rotate_right(11)
14
/ \
7 17
/ \
4 11
insert 54, 13, 12:
14
/ \
7 17
/ \ \
4 11 54
\
13
/
12
Node 11 doesn’t satisfy AVL.
rotate_right(13)
11
\
12
\
13
rotate_left(11)
14
/ \
7 17
/ \ \
4 12 54
/ \
11 13
insert 8:
14
/ \
7 17
/ \ \
4 12 54
/ \
11 13
/
8
Node 7 doesn’t satisfy AVL. There’s a zig-zag.
rotate_right(12)
14
/ \
7 17
/ \ \
4 11 54
/ \
8 12
\
13
left_rotate(7)
14
/ \
11 17
/ \ \
7 12 54
/ \ \
4 8 13
Given the following AVL Tree, delete the node with key 8. (This example has two nodes that end up violating the AVL property.)
8
/ \
5 10
/ \ / \
2 6 9 11
/ \ \ \
1 3 7 12
\
4
Solution:
Step 1: replace node 8 with node 9
9
/ \
5 10
/ \ \
2 6 11
/ \ \ \
1 3 7 12
\
4
Step 3: rotate 10 left
9
/ \
5 11
/ \ / \
2 6 10 12
/ \ \
1 3 7
\
4
Step 5: rotate 9 right
5
/ \
/ \
2 9
/ \ / \
1 3 6 11
\ \ / \
4 7 10 12
add or remove using the BST algorithm: O(log n)
number of iterations of the below “repeat”? O(log n) how much time per iteration (fix step): O(1) O(log n) * O(1) = O(log n)
total for add/remove and rebalance: O(log n) + O(log n) = O(log n)
Starting from the lowest changed node, repeat the following up to the root of the tree (because there can be several AVL violations).
if height(x.left) ≤ height(x.right) O(1)
if height(x.right.left) ≤ height(x.right.right)
let k = height(x.right.right)
x k+2 y ≤k+2
/ \ left_rotate(x) / \
≤k A y k+1 ===============> ≤k+1 x C k
/ \ / \
≤k B C k ≤k A B ≤k
if height(x.right.left) > height(x.right.right)
let k = height(x.right.left)
k+2 x y k+1
/ \ / \
k-1 A z k+1 R(z), L(x) k x z k
/ \ =============> / \ / \
k y D k-1 k-1 A B C D k-1
/ \
B C <k
if height(x.left) > height(x.right)
if height(x.left.left) < height(x.left.right) (note: strictly less!)
let k = height(x.left.right)
x k+2 z k+1
/ \ / \
k+1 y D k-1 L(y), R(x) k y x k
/ \ =============> / \ / \
k-1 A z k A B C D <k
/ \
B C <k
if height(x.left.left) ≥ height(x.left.right) (note: greater-equal!)
let k = height(l.left.left)
x k+2 y k+1
/ \ right_rotate(x) / \
k+1 y C k-1 ===============> k A x k+1
/ \ / \
k A B ≤k ≤k B C k-1
O(h) = O(log n)
insert n items: O(n log(n))
in-order traversal: O(n)
overall time complexity: O(n log(n))
priority queue: insert, delete, min
ordered set: insert, delete, min, max, next, previous