[PPT] - Binary Search Trees 1 October 2020 OSU CSE 1 Faster Searching PowerPoint Presentation

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Binary Search Trees

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The BinaryTree component family can

be used to arrange the labels on binary tree nodes in a variety of useful ways

A common arrangement of labels, which

supports searching that is much faster than linear search, is called a binary search tree (BST)

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Faster Searching

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BSTs Are Very General

BSTs may be used to search for items of

any type T for which one has defined a total preorder, i.e., a binary relation on T that is total, reflexive, and transitive

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BSTs Are Very General

BSTs may be used to search for items of

any type T for which one has defined a total preorder, i.e., a binary relation on T that is total, reflexive, and transitive

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A binary relation on T may be viewed as a set of ordered pairs of T,

r as a boolean-valued function R of

two parameters of type T that is true iff that pair is in the set.

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BSTs Are Very General

BSTs may be used to search for items of

any type T for which one has defined a total preorder, i.e., a binary relation on T that is total, reflexive, and transitive

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The binary relation R is total whenever: for all x, y: T (R(x, y) or R(y, x))

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BSTs Are Very General

BSTs may be used to search for items of

any type T for which one has defined a total preorder, i.e., a binary relation on T that is total, reflexive, and transitive

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The binary relation R is reflexive whenever: for all x: T (R(x, x))

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BSTs Are Very General

BSTs may be used to search for items of

any type T for which one has defined a total preorder, i.e., a binary relation on T that is total, reflexive, and transitive

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The binary relation R is transitive whenever: for all x, y, z: T (if R(x, y) and R(y, z) then R(x, z))

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Simplifications

For simplicity in the following illustrations,

we use only one kind of example:

– T = integer – The ordering is ≤

For simplicity (and because of how we will

use BSTs), we assume that no two nodes in a BST have the same labels

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SLIDE 9

Simplifications

For simplicity in the following illustrations,

we use only one kind of example:

– T = integer – The ordering is ≤

For simplicity (and because of how we will

use BSTs), we assume that no two nodes in a BST have the same labels

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Both these simplifications are inessential: BSTs are not limited to these situations!

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BST Arrangement Properties

A binary tree is a BST whenever the

arrangement of node labels satisfies these two properties:

1. For every node in the tree, if its label is x

and if y is a label in that node’s left subtree, then y < x

2. For every node in the tree, if its label is x

and if y is a label in that node’s right subtree, then y > x

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The Big Picture

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x

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The Big Picture

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x

Every label y in this tree satisfies y < x

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The Big Picture

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x

Every label y in this tree satisfies y > x

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And It’s So Everywhere

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x

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And It’s So Everywhere

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x

Every label y in this tree satisfies y < x

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And It’s So Everywhere

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x

Every label y in this tree satisfies y > x

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Examples of BSTs

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1 3 2 4 5 7 5 3 6 9

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Non-Examples of BSTs

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1 4 2 3 5 1 5 3 4 9 2

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Non-Examples of BSTs

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1 4 2 3 5 1 5 3 4 9 2

Property 1 is violated here.

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Non-Examples of BSTs

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1 4 2 3 5 1 5 3 4 9 2

Property 1 is violated here.

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Non-Examples of BSTs

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1 4 2 3 5 1 5 3 4 9 2

Property 2 is violated here.

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Searching for x

Suppose you are trying to find whether

any node in a BST t has the label x

There are only two cases to consider:

– t is empty – t is non-empty

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Searching for x

Suppose you are trying to find whether

any node in a BST t has the label x

There are only two cases to consider:

– t is empty – t is non-empty

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Easy: Report x is not in t.

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Searching for x

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r

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Searching for x

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r

Does x = r? If so, report that x is in t. If not ...

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Searching for x

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r

Is x < r? If so, report the result of searching for x in this tree. If not ...

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Searching for x

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r

Then it must be the case that x > r. Report the result of searching for x in this tree.

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Why It’s Faster Than Linear Search

You need to compare to the root of the

tree, and then (only if the root is not what you’re searching for) search either the left

r the right subtree—but not both

– Compare to linear search, where you might have to look at all the items, which would be equivalent to searching both subtrees

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Example: Searching for 5

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8

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Example: Searching for 5

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8 Does 5 = 8? No ...

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Example: Searching for 5

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8 Is 5 < 8? Yes, so report the result of searching for 5 in this tree.

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Recursion

Searching the left subtree at this point

simply involves making a recursive call to the method that searches a BST

Against our usual advice about recursion,

let’s trace into that call and see what happens

– Why? Because some people, e.g., interviewers, may expect you to understand BSTs without mentioning recursion/induction

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Example: Searching for 5

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3 8

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Example: Searching for 5

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3 8

Does 5 = 3? No ...

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Example: Searching for 5

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Is 5 < 3? No ...

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Example: Searching for 5

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3 8

Then it must be the case that 5 > 3. Report the result of searching for 5 in this tree.

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It’s Another Recursive Call

Let’s continue tracing into calls ...

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Example: Searching for 5

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6 3 8

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Example: Searching for 5

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6 3 8

Does 5 = 6? No ...

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Example: Searching for 5

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6 3 8

Is 5 < 6? Yes, so report the result of searching for 5 in the (empty) left subtree.

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The Recursion Stops Here

Remember, we already noted that when

searching for something in an empty tree, we can simply report it is not there

No new recursive call results

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Example: Searching for 5

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6 3 8

How many nodes did the algorithm visit, and compare labels to 5? At worst, how many could it be?

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Example: Searching for 5

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5 8

What about in this tree?

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Wait! How Can This Work?

With the BinaryTree components, there

are no methods to “move down the tree”

This is why recursion is crucial

– To search a subtree, you disassemble the

riginal tree, search in one of the subtrees,

and then (re)assemble it before returning the answer

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Refined Searching for x

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r

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Refined Searching for x

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r

Does x = r? If so, report that x is in t. If not ...

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Refined Searching for x

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r

Disassemble t into the root and its two subtrees.

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Refined Searching for x

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r

Is x < r? If so, remember the result of searching for x in this tree. If not ...

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Refined Searching for x

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r

Then it must be the case that x > r. Remember the result of searching for x in this tree.

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Refined Searching for x

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r

Before returning the result of the search, (re)assemble t from its parts.

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Inserting x

Suppose now you are trying to insert into a

BST t the label x (which we assume to be not already in t; remember that there are no duplicate labels in t)

There are only two cases to consider:

– t is empty – t is non-empty

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SLIDE 52

Inserting x

Suppose now you are trying to insert into a

BST t the label x (which we assume to be not already in t; remember that there are no duplicate labels in t)

There are only two cases to consider:

– t is empty – t is non-empty

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Easy: Make x the root

f the updated t.

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Inserting x

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r

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Inserting x

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r

There is no reason to ask whether x = r; why?

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Inserting x

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r

Is x < r? If so, insert x into this tree. If not ...

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Inserting x

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r

Then it must be the case that x > r. Insert x into this tree.

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Removing the Smallest

Suppose now you are trying to remove

from a BST t the smallest label (assuming that t is not empty)

There is only one case to consider:

– t is non-empty

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Removing the Smallest

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r

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Removing the Smallest

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r

Does the root have a non-empty left subtree?

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Removing the Smallest

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r

If so, remove the smallest label from this tree.

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Removing the Smallest

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r

If not, then r is the smallest label in t. Make the right subtree the new value of t, and return r.

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Removing x

Suppose now you are trying to remove

from a BST t the label x (which we assume to be in t)

There is only one case to consider:

– t is non-empty

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Removing x

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r

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Removing x

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r

Does x = r? If so, ouch! (Later ...) If not ...

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Removing x

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r

Is x < r? If so, remove x from this tree. If not ...

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Removing x

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r

Then it must be the case that x > r. Remove x from this tree.

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Removing x

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x

Back to the problematic case: x = r, i.e., we need to remove the root of t. What can we do?

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The Idea

To avoid restructuring the entire tree, we

can move to the root some label from further down in the tree that would not violate the BST arrangement properties

There are two good possibilities for the

label to be moved:

– The next-smaller label than x – The next-larger label than x

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Leverage Prior Work

We already know how to remove the

smallest label from a tree

So, it is easier to implement this strategy:

– Remove the smallest label from the right subtree (because this is the next-larger label after x in the original tree t) – Make that label the new root of t

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Leverage Prior Work

We already know how to remove the

smallest label from a tree

So, it is easier to implement this strategy:

– Remove the smallest label from the right subtree (because this is the next-larger label after x in the original tree t) – Make that label the new root of t

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But first, what if the right subtree is empty?

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Removing x

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x

If the right subtree is empty, then make the left subtree the new value of t.

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Removing x

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x s

If the right subtree is not empty, then replace x in the root with the smallest label from the right subtree.

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Resources

Big Java (4th ed), Section 16.5

– https://library.ohio-state.edu/record=b8540788~S7

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