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<!doctype html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>CS 2150: 05-trees slide set</title>
<meta name="description" content="A set of slides for a course on Program and Data Representation">
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<div class="reveal">
<!-- Any section element inside of this container is displayed as a slide -->
<div class="slides">
<section data-markdown><script type="text/template">
# CS 2150
### Program and Data Representation
<center><small><a href="http://www.cs.virginia.edu/~asb">Aaron Bloomfield</a> ([email protected])</small></center>
<center><small><a href="http://@github/uva-cs/pdr">@github</a> | <a href="index.html">↑</a> | <a href="daily-announcements.html?print-pdf"><img class="print" width="20" src="../slides/images/print-icon.png"></a></small></center>
## Trees
</script></section>
<section>
<h2>CS 2150 Roadmap</h2>
<table class="wide">
<tr><td colspan="3"><p class="center">Data Representation</p></td><td></td><td colspan="3"><p class="center">Program Representation</p></td></tr>
<tr>
<td class="top"><small> <br> <br>string<br> <br> <br> <br>int x[3]<br> <br> <br> <br>char x<br> <br> <br> <br>0x9cd0f0ad<br> <br> <br> <br>01101011</small></td>
<!-- image adapted from http://openclipart.org/detail/3677/arrow-left-right-by-torfnase -->
<td><img class="noborder" src="images/red-double-arrow.png" height="500" alt="vertical red double arrow"></td>
<td class="top"> <br>Objects<br> <br>Arrays<br> <br>Primitive types<br> <br>Addresses<br> <br>bits</td>
<td> </td>
<td class="top"><small> <br> <br>Java code<br> <br> <br>C++ code<br> <br> <br>C code<br> <br> <br>x86 code<br> <br> <br>IBCM<br> <br> <br>hexadecimal</small></td>
<!-- image adapted from http://openclipart.org/detail/3677/arrow-left-right-by-torfnase -->
<td><img class="noborder" src="images/green-double-arrow.png" height="500" alt="vertical green double arrow"></td>
<td class="top"> <br>High-level language<br> <br>Low-level language<br> <br>Assembly language<br> <br>Machine code</td>
</tr>
</table>
</section>
<section data-markdown><script type="text/template">
# Contents
[Introduction](#/introduction)
[Binary Search Trees](#/bsts)
[Expression Trees](#/exptrees)
[AVL Trees](#/avltrees)
[Recursion](#/recursion)
[Red-black Trees](#/redblacktrees)
[Splay Trees](#/splaytrees)
[Applications](#/applications)
</script></section>
<section>
<section id="introduction" data-markdown><script type="text/template">
# Introduction
</script></section>
<section data-markdown><script type="text/template">
## Data Structures
- If we have a good list implementation, do we need any other data structures?
- For computing: ***no***
- We can compute everything with just lists (actually even less). The underlying machine memory can be thought of as a list
- For thinking: ***yes***
- And for organizing the data
- Lists are a very limited way of thinking about problems
</script></section>
<section data-markdown><script type="text/template">
## List Limitations
- In a list, every element has direct relationships with only two others: the predecessor and the successor
- Access time: Θ(*n*)
- Goal: Θ(log *n*)
</script></section>
<section>
<h2><a href="http://en.wikipedia.org/wiki/File:Tree_of_life_SVG.svg">Complex Relationships: Phylogenetic Tree</a></h2>
<img alt="phylogenetic tree" class="stretch" src="images/05-trees/768px-Tree_of_life_SVG.svg.png" style="background-color:white">
</section>
<section>
<h2><a href="http://commons.wikimedia.org/wiki/File:IndoEuropeanTree.svg">Complex Relationships: Language Tree</a></h2>
<img alt="language tree" class="stretch" src="images/05-trees/IndoEuropeanTree.svg" style="background-color:white">
</section>
<section data-markdown><script type="text/template">
## List → Tree
- List: each element has relationships with up to 2 other elements:
- Binary Tree: each element has relationships with up to ***3*** other elements
- A tree is a special case of a list
</script></section>
<section data-markdown><script type="text/template">
## Tree Terms
- *Root*: a node with no parent; there can only be one root
- *Leaf*: a node with no children
- *Siblings*: two nodes with the same parent
- *Height* of a node: length of the ***longest*** path from that node to a leaf
- Thus, all leaves have height of zero
- *Height of a tree*: maximum depth of a node in that tree = height of the root
- *Depth* of a node: length of the path from the root to that node
</script></section>
<section>
<table class="transparent"><tr>
<td class="top" style="width:50%;"><a href="http://commons.wikimedia.org/wiki/File:IndoEuropeanTree.svg"><img src="images/05-trees/IndoEuropeanTree-part.svg" alt="language tree part" style="width:100%;background-color:white"></a></td>
<td class="top" style="width:50%;">
<h2>Tree Terms</h2>
<ul>
<li>Indo-European is the root</li>
<li>Nodes stacked on top of each other are siblings</li>
<li>English is a leaf, and has depth 5<ul class="small">
<li>Depth is the distance to the <i>root</i></li>
<li>English → Old English → Anglo-Frisian → West Germanic → Germaic → Indo-European</li></ul></li>
<li>Germanic is an internal node and has height 6<ul class="small">
<li>Height is the distance to the <i>furthest</i> leaf</li>
<li>Germanic → West Germanic → Low Franconian → West Low Franconian → Old Dutch → Dutch → Afrikaans</li></ul></li>
<li>The tree rooted at Germanic forms a sub-tree of the overall tree</li>
</ul>
</td></tr></table>
</section>
<section data-markdown><script type="text/template">
## More Tree Terms
- *Path*: sequence of nodes *n*<sub>1</sub>, *n*<sub>2</sub>, ..., *n*<sub>*k*</sub> such that *n*<sub>*i*</sub> is parent of *n*<sub>*i*+1</sub> for 1 ≤ *i* ≤ *k*
- *Length*: number of edges in the path
- *Internal path length*: sum of the depths of all the nodes
</script></section>
<section>
<h2>Other Examples of Trees</h2>
<table class="transparent"><tr><td class="top" style="width:60%">
<ul><li>Files and folders on a computer</li>
<li>Compilers: parse tree<pre><code class="avrasm">a = (b+c) * d;</code></pre></li>
<li>Genealogy trees<ul>
<li>These become complicated with some complex family relationships</li></ul></li>
<li>Lab 5: expression trees</li>
</ul>
</td><td class="top" style="width:40%"><img alt="expression tree 1" src="graphs/exp-tree-1.svg"></td>
</tr></table>
</section>
<section>
<h2>First child/next sibling</h2>
<table class="transparent"><tr>
<td class="middle"><pre><code class="avrasm">class TreeNode {
private:
string element;
TreeNode *firstChild;
TreeNode *nextSibling;
public:
// ...
}</code></pre></td>
<td><img alt="file tree 1" src="graphs/file-tree-1.svg"></td>
</tr></table>
</section>
<section data-markdown><script type="text/template">
## Traversals of trees
- Pre-order: `/ * + 1 2 - 3 4 * 5 6`
- In-order: `(((1+2) * (3-4)) / (5*6))`
- Parentheses are added to handle operator precedence issues
- Post-order: `1 2 + 3 4 - * 5 6 * /`
</script></section>
<section data-markdown><script type="text/template">
## Pre-order Traversal
- Pre-order: node first, then children (this is pseudo-code):
```
TreeNode::printTree() {
this.print();
for each child c of this:
c.printTree();
}
```
</script></section>
<section data-markdown><script type="text/template">
## In-order Traversal
- In-order: left node first, then self, then right node (C++):
```
void BST::print(BinaryNode *curNode) {
if (curNode != NULL) {
print(curNode->left);
cout << curNode->element << endl;
print(curNode->right);
}
}
```
</script></section>
<section data-markdown><script type="text/template">
## Post-order Traversal
- Post-order: children first, then node (partly C++, partly pseudo-code):
```
int TreeNode::numNodes(TreeNode *tnode) {
if ( tnode == NULL )
return 0;
else {
sum=0;
for each child c of tnode
sum += numNodes(c);
return 1 + sum;
}
}
```
</script></section>
<section data-markdown><script type="text/template">
## Traversals of trees (again)
- Pre-order: `/ * + 1 2 - 3 4 * 5 6`
- In-order: `(((1+2) * (3-4)) / (5*6))`
- Parentheses are added to handle operator precedence issues
- Post-order: `1 2 + 3 4 - * 5 6 * /`
</script></section>
</section>
<section>
<section id="bsts" data-markdown><script type="text/template">
# Binary Search Trees
</script></section>
<section>
<h2>Binary Trees</h2>
<table class="transparent"><tr>
<td class="middle" style="width:50%">
<p class="center">All nodes have at most 2 children</p>
<pre><code class="avrasm">class BinaryNode {
public:
// ...
private:
int element;
BinaryNode *left;
BinaryNode *right;
};</code></pre>
<img alt="bst-3" src="graphs/bst-3.svg"></td>
<td style="width:50%"><img alt="bst-1" src="graphs/bst-1.svg"></td>
</tr></table>
</section>
<section>
<h2>Binary Trees: diagram details</h2>
<table class="transparent"><tr>
<td class="top" style="width:60%">
<p>In reality, any child not shown is really a <code>NULL</code> pointer, as shown here; but these are generally omitted from the diagrams</p>
<img alt="bst-4" src="graphs/bst-4.svg"></td>
<td style="width:40%"><img alt="bst-2" src="graphs/bst-2.svg"></td>
</tr></table>
</section>
<section data-markdown><script type="text/template">
## Binary Search Trees (BST)
- Each node has a *key* value that can be compared
- Binary search tree property:
- For a given node, which we will call the *root*...
- Every node in left subtree has a key whose value is *less* than the root's key value, AND
- Every node in right subtree has a key whose value is *greater* than the root's key value
- We assume that duplicate values are not allowed
</script></section>
<section data-markdown><script type="text/template">
## BST: Example
</script></section>
<section data-markdown><script type="text/template">
## BST: Counter-example
</script></section>
<section data-markdown><script type="text/template">
## The difference
- Both binary trees and binary search trees have zero, one, or two children per node
- But a binary search tree is *sorted*
- However, most people, when they say "binary tree", really mean a "binary search tree"
- Note that we assume that we can *NOT* have duplicate elements in a BST
</script></section>
<section data-markdown><script type="text/template">
## BST: find
- Basic idea:
- Compare value to be found to key of the root of the tree
- If they are equal, then done
- If not equal, recurse depending on which half of tree the value should be in if it is in tree
- If you hit a `NULL` pointer, then you have "run off" the bottom of the tree, and the value is not in the tree
</script></section>
<section data-markdown><script type="text/template">
## BST: Find
- Trying to find 3 will go, from the root, left → left → right
- Trying to find 6 will go, from the root, right → left → left
- At that point, we have "run off" the bottom of the tree (via 7's left-child pointer, which is `NULL`), and thus the value is not in the tree
</script></section>
<section data-markdown><script type="text/template">
## BST: find
(no external source code)
```
BinaryNode * BST::find(int x, BinaryNode *curNode) {
// handle case where a NULL pointer could be passed
// curNode->right or curNode->left might be NULL
if (curNode == NULL) // we've "run" off the bottom
return NULL;
else if (x < curNode->element)
return find(x, curNode->left); // search left
else if (x > curNode->element)
return find(x, curNode->right); //search right
else
return curNode; // matched
}
```
</script></section>
<section>
<h2>BST: insert</h2>
<p>Do a find, and when we reach a <code>NULL</code> pointer, create a new node there</p>
<p>(no external source code)</p>
<pre><code>void BST::insert(int x, BinaryNode * & curNode) {
if (curNode==NULL)
curNode = new BinaryNode(x,NULL,NULL);
else if (x < curNode->element)
insert(x, curNode->left);
else if (x > curNode->element)
insert(x, curNode->right);
else
; // duplicate... do nothing
}
</code></pre>
</section>
<section data-markdown><script type="text/template">
## BST: findMax(), findMin()
To find the maximum element in BST, traverse down the right subtree links
Similarly down the left subttree links for `findMin()`
</script></section>
<section data-markdown><script type="text/template">
## BST: remove
- Disrupts the tree structure
- Basic idea:
- Find node to be removed
- Three cases:
- Node has no children
- Node has one child
- Node has two children
</script></section>
<section>
<h2>BST: remove: no children</h2>
<ul>
<li>Just remove the node (reclaiming memory), adjusting the parent pointer to <code>NULL</code><ul>
<li>In this case, 9's left child link is changed to <code>NULL</code></li></ul></li>
</ul>
<table class="transparent"><tr>
<td class="top"><img alt="bst-7" src="graphs/bst-7.svg"></td>
<td class="middle">→</td>
<td class="top"><img alt="bst-8" src="graphs/bst-8.svg"></td>
</tr></table>
</section>
<section>
<h2>BST: remove: one child</h2>
<ul>
<li>Adjust pointer of parent to point at child, and reclaim memory<ul>
<li>In this case, 4's left pointer is changed to point to 3</li></ul></li>
</ul>
<table class="transparent"><tr>
<td class="top"><img alt="bst-9" src="graphs/bst-9.svg"></td>
<td class="middle">→</td>
<td class="top"><img alt="bst-10" src="graphs/bst-10.svg"></td>
</tr></table>
</section>
<section>
<h2>BST: remove: two children</h2>
<ul>
<li>Replace node with successor, then remove successor from tree<ul>
<li>This requires running <code>findMin()</code> on the right sub-tree, and then removing that element</li>
<li>In this case, 5 is replaced by 7 (and the node that had 7 is removed)</li>
</ul></li></ul>
<table class="transparent"><tr>
<td class="top"><img alt="bst-11" src="graphs/bst-11.svg"></td>
<td class="middle">→</td>
<td class="top"><img alt="bst-12" src="graphs/bst-12.svg"></td>
</tr></table>
</section>
<section>
<h2>BST Height</h2>
<table class="transparent"><tr><td class="top" style="width:50%;">
<ul><li><i>n</i>-node BST: Worst case depth is <i>n</i>-1</li>
<li>This can easily happen if the data to be inserted is already sorted</li>
<li>Claim: The maximum number nodes in a binary tree of height <i>h</i> is 2<sup><i>h</i>+1</sup>-1</li></ul></td>
<td style="width:50%;"><img alt="bst-1" src="graphs/bst-1.svg"></td></tr></table>
</section>
<section data-markdown><script type="text/template">
## Proof by Induction on *h*
- Claim: max nodes in a binary tree of height *h* is 2<sup>*h*+1</sup>-1
- Or: *n* ≤ 2<sup>*h*+1</sup>-1
- For *h*=0: number of nodes is 2<sup>0+1</sup>-1 = 1
- Assume the claim is true for any tree of height *h*
- This would mean *n* ≤ 2<sup>*h*+2</sup>-1 for a tree of height *h*+1
- A tree of height *h*+1 can have 2 subtrees of height *h*; each subtree has 2<sup>*h*+1</sup>-1 nodes; add one for the root
- Thus, our new tree of height *h*+1 has:
- 2(2<sup>*h*+1</sup>-1)+1 = 2<sup>*h*+2</sup>-1 nodes
- If we put *h*+1 into our inductive hypothesis (instead of *h*), we get the same value; thus, it is proven
</script></section>
<section data-markdown><script type="text/template">
## Relationship between *h* and *n*
- Given *n* nodes and height *h*, then by the claim (proven on the previous slide): *n* ≤ 2<sup>*h*+1</sup>-1
- We can simplify:
- *n*+1 ≤ 2<sup>*h*+1</sup>
- log<sub>2</sub>(*n*+1) ≤ log<sub>2</sub>(2<sup>*h*+1</sup>)
- log<sub>2</sub>(*n*+1) ≤ *h*+1
- Thus *h* ≥ log<sub>2</sub>(*n*+1)-1
- This means that the "shortest" tree we can achieve for *n* nodes is proportional to the base-2 log of the height
</script></section>
<section data-markdown><script type="text/template">
## Perfect Binary Tree
- All leaves have the same depth
- And all nodes have zero or two children, but not one
- Number of leaves: 2<sup>*h*</sup>
- Number of nodes: 1 + 2 + 2<sup>2</sup> + 2<sup>3</sup> + ... + 2<sup>*h*</sup> = 2<sup>*h*+1</sup>-1
- Problem: a perfect binary only holds 2<sup>*h*+1</sup>-1 values
- So you can't have 5 values in a perfect binary tree!
</script></section>
</section>
<section>
<section id="exptrees" data-markdown><script type="text/template">
# Expression Trees
</script></section>
<section data-markdown><script type="text/template">
## Expression Trees
- A way to keep an internal representation of a mathematical equation
- This allows a computer to compute the value of each node
- You will have to do this for [lab 5](../labs/lab05/index.html)
- The lab expression tree only takes in integers as input; the more generalized version presented here includes variables as well
- See the [Wikipedia article on Expression trees](http://en.wikipedia.org/wiki/Expression_tree), specifically the [section on construction of expression trees](http://en.wikipedia.org/wiki/Expression_tree#Construction_of_an_Expression_Tree) (same example as this one)
- Uses a stack!
</script></section>
<section data-markdown><script type="text/template">
## Expression Tree Traversals
- Infix notation: `(a + ((b+c) * d))`
- For our purposes, we will always place parentheses around all operations, even if they aren't needed
- Postfix notation: `a b c + d * +`
- Prefix notation: `+ a * + b c d`
</script></section>
<section data-markdown><script type="text/template">
## Building an expression tree
- Given an expression in postfix notation, the algorithm to bulid an expression tree is somewhat similar to that to evaluate a postfix expression (from lab 3)
- The algorithm is to read tokens from input:
- If a number or a varaiable, push it on the stack
- If an operator, pop off two values, attach them as children of the operator, and push that back onto the stack
- A proper postfix expression will have only one value in the stack upon completion
</script></section>
<section data-markdown><script type="text/template">
## Example
Consider the postfix expression: a b + c d e + \* \*; the final expression tree is:
</script></section>
<section data-markdown><script type="text/template">
## Example, step 1
- Expression: <font color='red'>a</font> b + c d e + \* \*
- We read a, and push it onto the stack
</script></section>
<section data-markdown><script type="text/template">
## Example, step 2
- Expression: a <font color='red'>b</font> + c d e + \* \*
- We read b, and push it onto the stack
</script></section>
<section data-markdown><script type="text/template">
## Example, step 3
- Expression: a b <font color='red'>+</font> c d e + \* \*
- We read '+', pop off two values (a and b), put them as children of '+', and push the result back the stack
</script></section>
<section data-markdown><script type="text/template">
## Example, step 4
- Expression: a b + <font color='red'>c</font> d e + \* \*
- We read c, and push it onto the stack
</script></section>
<section data-markdown><script type="text/template">
## Example, step 5
- Expression: a b + c <font color='red'>d</font> e + \* \*
- We read d, and push it onto the stack
</script></section>
<section data-markdown><script type="text/template">
## Example, step 6
- Expression: a b + c d <font color='red'>e</font> + \* \*
- We read e, and push it onto the stack
</script></section>
<section data-markdown><script type="text/teplate">
## Example, step 7
- Expression: a b + c d e <font color='red'>+</font> \* \*
- We read '+', pop off two values (d and e), put them as children of '+', and push the result back the stack
</script></section>
<section data-markdown><script type="text/template">
## Example, step 8
- Expression: a b + c d e + <font color='red'>\*</font> \*
- We read '\*', pop off two values ('+' and c), put them as children of '\*', and push the result back the stack
</script></section>
<section data-markdown><script type="text/template">
## Example, step 9
- Expression: a b + c d e + \* <font color='red'>\*</font>
- We read '\*', pop off two values ('\*' and '+'), put them as children of '\*', and push the result back the stack
</script></section>
<section data-markdown><script type="text/template">
## The final result
</script></section>
</section>
<section>
<section id="avltrees" data-markdown><script type="text/template">
# AVL Trees
</script></section>
<section data-markdown><script type="text/template">
## Animation Tools
- A good AVL tree animation tool is [here](http://webdiis.unizar.es/asignaturas/EDA/AVLTree/avltree.html)
- We'll be using this website throughout this slide set
</script></section>
<section data-markdown><script type="text/template">
## AVL Trees
- Motivation: to ***guarantee*** Θ(log *n*) running time on find, insert, and remove
- Idea: Keep tree balanced after each operation
- Solution: AVL trees
- Named after the inventors, Adelson-Velskii and Landis
</script></section>
<section data-markdown><script type="text/template">
## AVL Tree Structure Property
For every node in the tree, the *height* of the left and right sub-trees differs at most by 1
</script></section>
<section data-markdown><script type="text/template">
## AVL Tree
</script></section>
<section data-markdown><script type="text/template">
## AVL balance factor
- Each node of a BST holds:
- The data
- Left and right child pointers
- Possibly a parent node pointer
- An AVL tree also holds a balance factor
- The height of the *right* subtree minus the height of the *left* subtree
- We could have it be left minus right, but the convention in this class is to always have it be right minus left
- Can be computed on the fly, as well, but that's VERY slow, and defeats the purpose of using AVL trees for speed
</script></section>
<section data-markdown><script type="text/template">
## AVL tree balance
- "Balanced" trees
- 0 means balanced
- 1 means the right subtree is one longer than the left subtree
- -1 means the left subtree is one longer than the right subtree
- "Unbalanced" trees
- A balance factor of -2 or 2
- We'll fix the tree
- Will we ever hit -3 or 3?
</script></section>
<section data-markdown><script type="text/template">
## AVL Tree, with balance factors
By definition, a BST is a valid AVL tree if the balance factor for ***EVERY*** node is -1, 0, or 1
</script></section>
<section data-markdown><script type="text/template">
## Not an AVL Tree
Not balanced: height difference greater than 1
</script></section>
<section data-markdown><script type="text/template">
## AVL Trees: find, insert
- find: same as BST find
- insert: same as BST insert, except might need to "fix" the AVL tree after the insert (via rotations)
- Runtime analysis:
- Θ(*d*), where *d* is the depth of the node being found/inserted
- What is the maximum height of an n-node AVL tree?
</script></section>
<section data-markdown><script type="text/template">
## AVL tree operations
- Perform the operation (insert, delete)
- Move back up to the root, updating the balance factors
- Why only those nodes?
- Because those are the only ones who have had their subtrees altered
- Do tree rotations where the balance factors are 2 or -2
</script></section>
<section data-markdown><script type="text/template">
## How many times to "fix" the tree?
- Any single insert will only modify the balance factor by one
- So we fix the lowest off-balance nodes
- Then everything above it is then balanced
- This means that we will have to only look at the bottom two unbalanced nodes
</script></section>
<section data-markdown><script type="text/template">
## AVL insert
- Let *x* be the *deepest* node where imbalance occurs
- Four cases where the insert happened:
1. In the left subtree of the left child of x
2. In the right subtree of the left child of x
3. In the left subtree of the right child of x
4. In the right subtree of the right child of x
- Cases 1 & 4: perform a single rotation
- Cases 2 & 3: perform a double rotation
</script></section>
<section>
<h2>AVL single right rotation</h2>
<table class="transparent"><tr>
<td class="top"><img alt="avl-tree-18" src="graphs/avl-tree-18.svg"></td>
<td class="middle">→</td>
<td class="top"><img alt="avl-tree-19" src="graphs/avl-tree-19.svg"></td>
</tr></table>
<ul>
<li>The node just inserted was node 1 (blue)</li>
<li>The <i><b>lowest</b></i> node, immediately after the insert, with an imbalance is node 3 (red)</li>
<li>Because node 1 is in the "left subtree of the left child" of node 3, this means we need to perform a single right rotation</li>
</ul>
</section>
<section>
<h2>AVL single left rotation</h2>
<table class="transparent"><tr>
<td class="top"><img alt="avl-tree-20" src="graphs/avl-tree-20.svg"></td>
<td class="middle">→</td>
<td class="top"><img alt="avl-tree-19" src="graphs/avl-tree-19.svg"></td>
</tr></table>
<ul>
<li>The node just inserted was node 3 (red)</li>
<li>The <i><b>lowest</b></i> node, immediately after the insert, with an imbalance is node 1 (blue)</li>
<li>Because node 3 is in the "right subtree of the right child" of node 1, this means we need to perform a single left rotation</li>
</ul>
</section>
<section>
<h2>A side-effect of tree rotations</h2>
<table class="transparent"><tr>
<td class="top"><img alt="avl-tree-18" src="graphs/avl-tree-18.svg"></td>
<td class="middle">→</td>
<td class="top"><img alt="avl-tree-19" src="graphs/avl-tree-19.svg"></td>
</tr></table>
<ul>
<li>This is the single right rotation</li>
<li>Note that at least one node moves "up" (depth decreases)<ul>
<li>In this case, nodes 1 and 2 both move up</li></ul></li>
<li>And at least one node moves "down" (depth increases)<ul>
<li>In this case, node 3 moves down</li></ul></li>
<li>Similarly for a left rotation</li>
</ul>
</section>
<section>
<h2>AVL single right rotation: before & after</h2>
<table class="transparent"><tr>
<td><img alt="avl tree 4" src="graphs/avl-tree-4.svg" width="350"></td>
<td class="middle">→</td>
<td class="top"><img alt="avl tree 5" src="graphs/avl-tree-5.svg" width="350"></td>
</tr></table>
<ul>
<li>Node 1 (red) is what is being inserted</li>
<li>The <i><b>lowest</b></i> node with an imbalance is node 5 (balance: -2)</li>
<li>Because the insert was in 5's "left subtree of the left child", we perform a single right rotation on 5 (and its left child, 3)</li>
</ul>
</section>
<section>
<h2>AVL single right rotation: before & after</h2>
<table class="transparent"><tr>
<td><img alt="avl tree 4" src="graphs/avl-tree-4.svg" width="350"></td>
<td class="middle">→</td>
<td class="top"><img alt="avl tree 5" src="graphs/avl-tree-5.svg" width="350"></td>
</tr></table>
<ul>
<li>From the previous slide, we know we perform a single right rotation on 5 (and its left child, 3)</li>
<li>Thus, the two blue nodes are the 'pivots' of the rotation</li>
<li>Note that node 4 changes parents (from 3's right to 5's left)</li>
</ul>
</section>
<section>
<h2>AVL single right rotation: general case</h2>
<table class="transparent"><tr>
<td><img alt="avl tree 6" src="graphs/avl-tree-6.svg"></td>
<td class="middle">→</td>
<td><img alt="avl tree 7" src="graphs/avl-tree-7.svg"></td>
</tr></table>
<p class="center">\( X < b < Y < a < Z \)</p>
<p class="center">The insert is into sub-tree X, increasing its height to <i>h</i>+1</p>
<p class="center">Notice how sub-tree Y changes parent</p>
</section>
<section data-markdown><script type="text/template">
## Right and left rotations
Note that the trees shown are not necessarily AVL trees, but the rotations are correct
[](http://en.wikipedia.org/wiki/File:Tree_rotation.png)
</script></section>
<section>
<h2>Cases 2 & 3: attempt a single rotation</h2>
<table class="transparent"><tr>
<td><img alt="avl tree 8" src="graphs/avl-tree-8.svg"></td>
<td class="middle">→</td>
<td><img alt="avl tree 9" src="graphs/avl-tree-9.svg"></td>
</tr></table>
<p class="center">\( X < b < Y < a < Z \)</p>
<p class="center">The insert is into sub-tree Y, increasing its height to <i>h</i>+1</p>
<p class="center">Failure! b's left subtree has height <i>h</i>+1; right is <i>h</i>+3</p>
</section>
<section>
<h2>Double rotation</h2>
<table class="transparent"><tr>
<td class="top" style="width:60%;">
<ul><li>Node 5 (red) was just inserted</li>
<li>The <i><b>lowest</b></i> node with an imbalance is node 8 (balance factor: -2)<ul>
<li>When discussing these rotations, we will call this the "parent" node</li></ul></li>
<li>Because the insert happened in 8's
"right subtree of the left child", we perform a <i>double</i> rotation</li>
<li>This consists of a single <i>left</i> rotation on the "child" (node 4), followed by a single <i>right</i> rotation on the "parent" (node 8)</li>
</ul></td>
<td class="top" style="width:40%;"><img alt="avl tree 14" src="graphs/avl-tree-14.svg" width="400"></td>
</tr>
<tr><td colspan="2" style="text-align:left;"><ul><li>Note that the two rotations are in different directions!</li></ul></td></tr>
</table>
</section>
<section>
<h2>Double rotation, step 1</h2>
<table class="transparent"><tr>
<td class="top"><img alt="avl tree 10" src="graphs/avl-tree-10.svg" width="400"></td>
<td class="middle">→</td>
<td class="top"><img alt="avl tree 11" src="graphs/avl-tree-11.svg" width="510"></td>
</tr></table>
<p>This is the single left rotation on the "child". The red node is what was inserted; the blue nodes are the 'pivots' of this single left rotation.</p>
</section>
<section>
<h2>Double rotation, step 2</h2>
<table class="transparent"><tr>
<td class="top"><img alt="avl tree 12" src="graphs/avl-tree-12.svg"></td>
<td class="middle">→</td>
<td class="top"><img alt="avl tree 13" src="graphs/avl-tree-13.svg"></td>
</tr></table>
<p>This is the single right rotation on the "parent". The red node is what was inserted; the green nodes are the 'pivots' of this single right rotation.</p>
</section>
<section>
<h2>AVL double rotation: before & after</h2>
<table class="transparent"><tr>
<td class="top"><img alt="avl tree 14" src="graphs/avl-tree-14.svg"></td>
<td class="middle">→</td>
<td class="top"><img alt="avl tree 15" src="graphs/avl-tree-15.svg"></td>
</tr></table>
<p class="center">The red node is what was inserted</p>
</section>
<section>
<h2>AVL double rotation: general case</h2>
<table class="transparent"><tr>
<td><img alt="avl tree 16" src="graphs/avl-tree-16.svg" width="275"></td>
<td class="middle">
<table><tr><td>
<table><tr><td class="middle">→</td><td><img alt="avl tree 17" src="graphs/avl-tree-17.svg"></td></tr></table>
</td></tr><tr><td> </td></tr><tr><td>\( W < b < X < c < Y < a < Z \)</td></tr><tr><td> </td></tr><tr><td>The insert happens into X</td></tr>
<tr><td>Notice sub-trees X <i>and</i> Y change parents</td></tr></table>
</td>
</tr></table>
</section>
<section data-markdown><script type="text/template">
## Ack! Terminology
- Some people will state a 'double left rotation'
- But is that a left-right? Or a right-left?
- We'll call them 'double left-right' and 'double right-left', which specifies the order to perform the operation on the child and then the parent
</script></section>
<section data-markdown><script type="text/template">
## AVL insert, again
- Let *x* be the *deepest* node where imbalance occurs
- Four cases where the insert happened:
1. In the left subtree of the left child of x
2. In the right subtree of the left child of x
3. In the left subtree of the right child of x
4. In the right subtree of the right child of x
- Cases 1 & 4: perform a single rotation
- Cases 2 & 3: perform a double rotation
</script></section>
<section>
<h2>Algorithmic determination of rotation</h2>
<table class="transparent">
<tr>
<td class="top"><img alt="avl tree 21" src="graphs/avl-tree-21.svg"></td>
<td class="top"><img alt="avl tree 22" src="graphs/avl-tree-22.svg"></td>
<td class="top"><img alt="avl tree 23" src="graphs/avl-tree-23.svg"></td>
<td class="top"><img alt="avl tree 24" src="graphs/avl-tree-24.svg"></td>
</tr>
<tr><td>left-left<br>case</td><td>right-right<br>case</td><td>left-right<br>case</td><td>right-left<br>case</td></tr>
<tr><td> </td><td> </td><td> </td><td> </td></tr>
</table>
<ul>
<li>Given the lowest unbalanced node, and the child in the direction of the insert, compare the balance factors</li>
<li>-2/+1 means a double left-right, +2/+1 means a single left, etc.</li>
</ul>
</section>
<section>
<h2><a href="http://en.wikipedia.org/wiki/File:Tree_Rebalancing.gif">All the tree rotations</a></h2>
<img class="stretch" alt="tree rotations" src="images/05-trees/1024px-Tree_Rebalancing.gif">
</section>
<section data-markdown><script type="text/template">
## AVL Tree: Runtime Analysis
- Find: Θ(log *n*) time: height of tree is always Θ(log *n*)
- Insert: Θ(log *n*) time: find() takes Θ(log *n*), then may have to visit every node on the path back up to root to perform up to 2 single rotations
- Remove: Θ(log *n*): left as an exercise
- Print: Θ(*n*): no matter the data structure, it will still take *n* steps to print *n* elements
</script></section>
</section>
<section>
<section id="recursion" data-markdown><script type="text/template">
# Recursion
</script></section>
<section data-markdown><script type="text/template">
## Sum the numbers from 1 to n
- We can do it iteratively:
```
int sum (int n) {
int s = 0;
for ( int i = 1; i <= n; i++ )
s += i;
return s;
}
```
</script></section>
<section data-markdown><script type="text/template">
## Sum the numbers from 1 to n
- We can do it *recursively*:
```
int sum (int n) {
return n + sum (n-1);
}
```
What problems arise?
</script></section>
<section data-markdown><script type="text/template">
## Recursion
- Recursion always needs three things to work:
- A way to make the problem *simpler* or smaller
- A way to *detect* when it should terminate or end
- A way to *terminate* or end
- The last example didn't end (among other issues)
</script></section>
<section data-markdown><script type="text/template">
## Sum the numbers from 1 to n
- We can do it *recursively*, but right this time:
```
unsigned int sum (unsigned int n) {
if ( n == 0 )
return 0;
else
return n + sum (n-1);
}
```
- Making the problem simpler is the `sum(n-1)`
- Detecting when to end is the `if (n==0)`
- Termination is the `return 0`
</script></section>
<section data-markdown><script type="text/template">
## Factorial via recursion
- We can do factorial recursively also:
```
unsigned long fact (unsigned long n) {
if ( n == 0 )
return 1;
else
return n * fact (n-1);
}
```
- Making the problem simpler is the `fact(n-1)`
- Detecting when to end is the `if (n==0)`
- Termination is the `return 1`
</script></section>
<section data-markdown><script type="text/template">
## Recursion: pros and cons
- Pros
- It's a more natural way to think of the problem, as you only focus on that one "instance"
- Some problems work well with recursion, but not with iteration (such as tree traversals)
- Cons
- Invoking a subroutine at each step slows performance
- We'll see a way to solve that shortly...
</script></section>
<section data-markdown><script type="text/template">
## Recursion: Fibonacci sequence
```
unsigned int fib (unsigned int n) {
if ( n <= 1 )