Overview:
Review BST Remove
Segment Intersection Project
Balanced Trees: AVL Trees
remove method of BinarySearchTreeBook 4.3.4.
| |
z=o A
\ ==>
A
| |
z=o A
/ ==>
A
|
z=o
/ \
A B
The main idea is to replace z with the node after z, which is the first node y in subtree B. We then recursively delete y from B.
Solution for remove() (similar to book Figure 4.25):
public void remove(K key) {
root = remove_helper(root, key);
}
private Node remove_helper(Node<K> curr, K key) {
if (curr == null) {
return null;
} else if (lessThan.test(key, curr.data)) { // remove in left subtree
curr.left = remove_helper(curr.left, key);
return curr;
} else if (lessThan.test(curr.data, key)) { // remove in right subtree
curr.right = remove_helper(curr.right, key);
return curr;
} else { // remove this node
if (curr.left == null) {
return curr.right;
} else if (curr.right == null) {
return curr.left;
} else { // two children, replace with first of right subtree
Node<K> min = curr.right.first();
curr.data = min.data;
curr.right = remove_helper(curr.right, min.data);
return curr;
}
}
}
What is the time complexity? $O(h)$, where $h$ is the height.
Given a set of n segments, are their any pairs that intersect?
Suppose we have a routine for testing whether 2 segments intersect.
Simplifications:
Brute force: test all combinations O(n²)
A better algorithm: Line Sweep
We’ll need fast membership testing, insert, remove, and next/previous.
Recall that search time is O(h), where h is the height of the tree.
Definition of height
int compute_height(Node n) {
if (n == null) {
return -1;
} else {
int hl = compute_height(n.left);
int hr = compute_height(n.right);
return 1 + Math.max(hl, hr);
}
}
Example tree with heights in brackets:
41[3]
/ \
20[2] 65[1]
/ \ /
11[0] 29[1] 50[0]
/
26[0]
The problem of unbalanced trees
o
\
o
\
o
\
o
\
o
\
o
\
o
height = n
vs.
o
/ \
/ \
o o
/ \ / \
o o o o
height = log(n)
Definition A tree is balanced if its height is O(log n) where n is the number of nodes in the tree.
Equivalently, the number of nodes is Ω(2ʰ) where h is the height.
Definition The AVL Invariant: the height of two child subtrees may only differ by 1.
Examples of trees that are AVL:
o o o o
/ / \ / \ / \
o o o o o o o
/ / \
o o o
Examples of trees that are not AVL:
o o o
/ \ / \
o o o o
\ \ / \
o o o o
\
o
Red-black trees are an alternative: AVL is faster on lookup than red-black trees but slower on insertion and removal because AVL is more rigidly balanced.
Let N(h) represent the minimum number of nodes in an AVL tree of height h. (The least-balanced possible scenario.)
N(h) = N(h-1) + N(h-2) + 1
We want to show that
h ≲ log₂ N(h)
To simplify, we have
N(h-2) + N(h-2) + 1 < N(h-1) + N(h-2) + 1 = N(h)
2·N(h-2) + 1 < N(h)
2·N(h-2) < N(h)
= 2·2·N(h-4)
= 2·2·2·N(h-6)
...
= 2^(h/2) < N(h)
Take the log of both sides
log₂ 2^(h/2) < log₂ N(h)
(log₂ Aᴮ = B log₂ A)
h/2 · log₂ 2 < log₂ N(h)
(log₂ 2 = 1, i.e. 2¹ = 2)
h/2 · 1 < log₂ N(h)
(multiply both side by 2)
h < 2 · log₂ N(h)
so we have
h ≲ log₂ N(h)
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
From the changed node on up (there can be several AVL violations)
if height(x.left) ≤ height(x.right)
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