Shreyas’ Notes

# COMP 382

## BSTs §

Assumption: keys are linearly ordered.

Let $x$ and $y$ be nodes in the tree.

• If node $y$ is in the left subtree of $x$, then $y_{key} \leq x_{key}$
• If node $y$ is in the right subtree of $x$, then $y_{key} \geq x_{key}$

Every subtree of a BST is a BST.

### Traversal §

in_order(x):
if x != null:
in_order(x.left)
print(x.key)
in_order(x.right)

• preorder: root, left, right
• inorder: left, root, right
• postorder: left, right, root

Running time: $\Theta(n)$ where $n$ is the tree size

tree_search(x, k):
if x == null or k == x.key
return x

if k < x.key:
return tree_search(x.left, k)

return tree_serach(x.right, k)


$O(h)$, where $h$ is the height of the tree. $h$ is

tree_search(x, k):
while x != null and k != x.key:
if k < x.key:
x = x.left
else:
x = x.right

return x


### Successor §

Pre-condition: x != null

tree_successor(x):
if x.right != null:
return tree_min(x)

y = x.p
while y != null and x == y.right:
x = y
y = y.p
return y


## Red-Black Trees §

1. every node is either red or black
2. the root is black
3. every leaf (NIL) is black
4. the children of a red node must be black
5. for every node $u$, all simple paths from $u$ to descendent leaves have the same number of black nodes.

properties 4 and 5 together ensure that the tree is “roughly balanced”.

## AVL Trees §

Binary search trees that are balanced.

For every node $x$ with a left subtree $l$ and a right subtree $r$, $|\textrm{height}(r) - \textrm{height}(l)| \leq 1$. The heights of subtrees $l$ and $r$ differ by at most 1.

Balance factor $\textrm{balance}(x) = \textrm{height}(x.r) - \textrm{height}(x.l)$

## B-Trees §

Assumption: each node is stored on a separate disk page.

The branching factor of B trees can be large.

A node with $n$ keys has $n + 1$ children.

A rooted tree:

• every node x has attributes
• number of keys stored x.n
• the keys x.key[1] <= x.key[2] <= ... <= x.key[x.n] (ordered)
• boolean value x.leaf
• every internal node x has x.n + 1 pointers x.c[1], x.c[2], …, x.c[x.n + 1] to its children
• the keys separate the ranges of keys for subtrees
• all leaves have the same depth, which is equal to the height h of the tree
• a b-tree has a parameter $t \geq 2$ called the minimum degree
• each node (except root) has at least $t - 1$ keys. each internal node (except root) has at least $t$ children.
• each node has at msot $2t - 1$ keys

Consider a non-empty B-tree $T$ with minimum degree $t$. The height of $T$ is $h$.

# assumption: x.node has already been read from disk k.key
b_tree_search(x, k):
i = 1
while i <= x.n and k > x.key[i]:
i++

if i <= x.n and k == x.key[i]:
return (x, i)
else if x.leaf:
return null
else:
return b_tree_search(x.c[i], k)


$O(h)$ reads from disk and $O(t \cdot h)$ running time.

## Hash Tables §

Data structure that stores keys with additional satellite data

### The universe is small §

Direct address table: array T[0..m-1].

• If key k is not present, T[k] = NIL
• If key k is present and has value val, T[k] = val

### The universe is large; the set $k$ of present keys is small §

Memory: $\Theta(|k|)$

Search: $O(1)$ time in the average case

Components:

• table T[0..m-1] where typically m \llt |U|
• hash function $h : U \rightarrow \{0, 1, \cdots, m - 1\}$
• key $k$ hashes to the slot $h(k)$

values stored in T[h(k)]. collisions unavoidable since $m < |U|$.

chaining: elements that hash to the same value are placed into a linked list.

• insert(T, x): insert x at the head of list T[h(x.key)]
• search(T, k): search for k in the list T[h(k)]
• delete(T, x): delete x from the list T[h(x.key)]

worst case for search: all $n$ keys hash to the same slot. searching will be $\Theta(n)$.

[Assumption: simple, uniform hashing]

We insert keys $k_1, k_2, \cdots, k_n$ into a hash table with $m$ slots

$P(h(i) = j) = \frac{1}{m}$

$n_j$ length of the list T[j]