(Binary tree)

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Data Structure - Binary Tree

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Binary Trees

2

Parts of a binary tree

A binary tree is composed of zero or more nodes Each node contains:

A value (some sort of data item) A reference or pointer to a left child (may be null), and A reference or pointer to a right child (may be null)

A binary tree may be empty (contain no nodes) If not empty, a binary tree has a root node

Every node in the binary tree is reachable from the root node by a unique path

A node with neither a left child nor a right child is called a leaf In some binary trees, only the leaves contain a value

3

Picture of a binary tree

a

b c

d e

g h i

l

f

j k

4

Size and depth

The size of a binary tree is the number of nodes in it This tree has size 12

The depth of a node is its distance from the root

a is at depth zero e is at depth 2

The depth of a binary tree is the depth of its deepest node This tree has depth 4

a

b c

d e f

g h i j k

l

5

Balance

A binary tree is balanced if every level above the lowest is “full” (contains 2n nodes)

In most applications, a reasonably balanced binary tree is desirable

a

b c

d e f g

h i j

A balanced binary tree

a

b

c

d

e

f

g h

i jAn unbalanced binary tree

6

Binary search in an array

Look at array location (lo + hi)/2

2 3 5 7 11 13 17 0 1 2 3 4 5 6

Searching for 5:(0+6)/2 = 3

hi = 2;(0 + 2)/2 = 1 lo = 2;

(2+2)/2=27

3 13

2 5 11 17

Using a binary search tree

7

Tree traversals

A binary tree is defined recursively: it consists of a root, a left subtree, and a right subtree

To traverse (or walk) the binary tree is to visit each node in the binary tree exactly once

Tree traversals are naturally recursive Since a binary tree has three “parts,” there are six possible

ways to traverse the binary tree: root, left, right left, root, right left, right, root

root, right, left right, root, left right, left, root

8

Preorder traversal

In preorder, the root is visited first Here’s a preorder traversal to print out all the

elements in the binary tree:

public void preorderPrint(BinaryTree bt) { if (bt == null) return; System.out.println(bt.value); preorderPrint(bt.leftChild); preorderPrint(bt.rightChild);}

9

Inorder traversal

In inorder, the root is visited in the middle Here’s an inorder traversal to print out all the

elements in the binary tree:

public void inorderPrint(BinaryTree bt) { if (bt == null) return; inorderPrint(bt.leftChild); System.out.println(bt.value); inorderPrint(bt.rightChild);}

10

Postorder traversal

In postorder, the root is visited last Here’s a postorder traversal to print out all the

elements in the binary tree:

public void postorderPrint(BinaryTree bt) { if (bt == null) return; postorderPrint(bt.leftChild); postorderPrint(bt.rightChild); System.out.println(bt.value);}

11

Tree traversals using “flags” The order in which the nodes are visited during a tree

traversal can be easily determined by imagining there is a “flag” attached to each node, as follows:

To traverse the tree, collect the flags:

preorder inorder postorder

A

B C

D E F G

A

B C

D E F G

A

B C

D E F G

A B D E C F G D B E A F C G D E B F G C A

12

Copying a binary tree

In postorder, the root is visited last Here’s a postorder traversal to make a complete

copy of a given binary tree:

public BinaryTree copyTree(BinaryTree bt) { if (bt == null) return null; BinaryTree left = copyTree(bt.leftChild); BinaryTree right = copyTree(bt.rightChild); return new BinaryTree(bt.value, left, right);}

13

Other traversals

The other traversals are the reverse of these three standard ones That is, the right subtree is traversed before the left subtree

is traversed Reverse preorder: root, right subtree, left subtree Reverse inorder: right subtree, root, left subtree Reverse postorder: right subtree, left subtree, root

14

Tree searches

A tree search starts at the root and explores nodes from there, looking for a goal node (a node that satisfies certain conditions, depending on the problem)

For some problems, any goal node is acceptable (N or J); for other problems, you want a minimum-depth goal node, that is, a goal node nearest the root (only J)

L M N O P

G

Q

H JI K

FED

B C

A

Goal nodes

15

Depth-first searching

A depth-first search (DFS) explores a path all the way to a leaf before backtracking and exploring another path

For example, after searching A, then B, then D, the search backtracks and tries another path from B

Node are explored in the order A B D E H L M N I O P C F G J K Q

N will be found before JL M N O P

G

Q

H JI K

FED

B C

A

How to do depth-first searching

Put the root node on a stack;while (stack is not empty) { remove a node from the stack; if (node is a goal node) return success; put all children of node onto the stack;}return failure;

At each step, the stack contains some nodes from each of a number of levels The size of stack that is required depends on the

branching factor b While searching level n, the stack contains

approximately b*n nodes When this method succeeds, it doesn’t give the path

Depth-First Search

A

B C

D E F G

A B D E C F GA B D E C F G

Stack:

Current:

Depth-First Search

A

B C

D E F G

Undiscovered

Marked

Finished

Active

B newly discovered

Stack:

(A, C); (A, B)

Current:

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

A alreadymarked

Visited B

Stack:

(B, E); (B, D); (B, A)(A, C); (A, B)

20

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

A alreadymarked

Visited B

AA

Stack:

(B, E); (B, D); (B, A)(A, C); (A, B)

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

D newly discovered

AA

Stack:

(B, E); (B, D); (B, A)(A, C); (A, B)

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

Visited D

A BA B

Stack:

(B, E); (B, D); (B, A)(A, C); (A, B)

(B, D)

Marked B

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

Finished D

A BA B

Stack:

(B, E); (B, D); (B, A)(A, C); (A, B)

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

E newly discovered

Stack:

(B, E); (B, D); (B, A)(A, C); (A, B)

A B DA B D

Backtrack B

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

Visited E

Stack:

(B, E); (B, D); (B, A)(A, C); (A, B)

A B DA B D

(B, E)

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

Finished E

Stack:

(B, E); (B, D); (B, A)(A, C); (A, B)

A B DA B D

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

Stack:

(B, E); (B, D); (B, A)(A, C); (A, B)

A B D EA B D E

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

Stack:

(A, C); (A, B)

A B D EA B D E

Backtrack B

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

Stack:

(A, C); (A, B)

A B D EA B D E

Backtrack A

Finished BC newly discovered

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

Stack:

(C, G); (C, F); (C, A)

A B D EA B D E

Visited C

(A, C); (A, B)

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

Stack:

(C, G); (C, F); (C, A)

Marked C

(A, C); (A, B)

F newly discovered

A B D E CA B D E C

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

Stack:

(C, G); (C, F); (C, A)(A, C); (A, B)

Visited F

A B D E CA B D E C

(F, C)

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

Stack:

(C, G); (C, F); (C, A)(A, C); (A, B)

Finished F

A B D E CA B D E C

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

Stack:

(C, G); (C, F); (C, A)(A, C); (A, B)

Backtrack C

G newly discovered

A B D E C FA B D E C F

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

Stack:

(C, G); (C, F); (C, A)(A, C); (A, B)

Marked C

Visited G

A B D E C FA B D E C F

(G, C)

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

Stack:

(C, G); (C, F); (C, A)(A, C); (A, B)

Backtrack C

Finished G

A B D E C F GA B D E C F G

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

Stack:

(A, C); (A, B)

Finished C

A B D E C F GA B D E C F G

Backtrack A

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active

Stack:

A B D E C F GA B D E C F G

Finished A

Depth-First Search

A

B C

D E F G

Current:Undiscovered

Marked

Finished

Active A B D E C F GA B D E C F G

40

Breadth-first searching

A breadth-first search (BFS) explores nodes nearest the root before exploring nodes further away

For example, after searching A, then B, then C, the search proceeds with D, E, F, G

Node are explored in the order A B C D E F G H I J K L M N O P Q

J will be found before NL M N O P

G

Q

H JI K

FED

B C

A

How to do breadth-first searching

Put the root node on a queue;while (queue is not empty) { remove a node from the queue; if (node is a goal node) return success; put all children of node onto the queue;}return failure;

Just before starting to explore level n, the queue holds all the nodes at level n-1

In a typical tree, the number of nodes at each level increases exponentially with the depth

Memory requirements may be infeasible When this method succeeds, it doesn’t give the path There is no “recursive” breadth-first search equivalent to

recursive depth-first search

Breadth-First Search

A

B C

D E F G

A B C D E F GA B C D E F G

Queue:

Current:

Breadth-First Search

A

B C

D E F G

Queue:

Current:

A

Breadth-First Search

A

B C

D E F G

Queue:

Current:

A

A

Breadth-First Search

A

B C

D E F G

Queue:

Current:A

AA

Breadth-First Search

A

B C

D E F G

Queue:

Current:

B

A

AA

Breadth-First Search

A

B C

D E F G

Queue:

Current:

CB

A

AA

Breadth-First Search

A

B C

D E F G

Queue:

Current:B

AA

CB

Breadth-First Search

A

B C

D E F G

Queue:

Current:B

C

A BA B

Breadth-First Search

A

B C

D E F G

Queue:

Current:

DC

B

A BA B

Breadth-First Search

A

B C

D E F G

Queue:

Current:

EDC

B

A BA B

Breadth-First Search

A

B C

D E F G

Queue:

Current:C

EDC

A BA B

Breadth-First Search

A

B C

D E F G

Queue:

Current:C

A B CA B C

ED

Breadth-First Search

A

B C

D E F G

Queue:

Current:C

A B CA B C

FED

Breadth-First Search

A

B C

D E F G

Queue:

Current:C

A B CA B C

GFED

Breadth-First Search

A

B C

D E F G

Queue:

Current:D

A B CA B C

GFED

A B C DA B C D

Breadth-First Search

A

B C

D E F G

Queue:

Current:D

GFE

Breadth-First Search

A

B C

D E F G

A B C DA B C D

Queue:

Current:E

GFE

Breadth-First Search

A

B C

D E F G

Queue:

Current:F

G

A B C D E FA B C D E F

Breadth-First Search

A

B C

D E F G

Queue:

Current:G

G

A B C D E FA B C D E F

Breadth-First Search

A

B C

D E F G

Queue:

Current:G

A B C D E F GA B C D E F G

Breadth-First Search

A

B C

D E F G

A B C D E F GA B C D E F G

63

Depth- vs. breadth-first searching

When a breadth-first search succeeds, it finds a minimum-depth (nearest the root) goal node

When a depth-first search succeeds, the found goal node is not necessarily minimum depth

For a large tree, breadth-first search memory requirements may be excessive

For a large tree, a depth-first search may take an excessively long time to find even a very nearby goal node

How can we combine the advantages (and avoid the disadvantages) of these two search techniques?

findMin/ findMax Return the node containing the smallest element in the tree Start at the root and go left as long as there is a left child.

The stopping point is the smallest element

Similarly for findMax Time complexity = O(height of the tree)

insert

Proceed down the tree as you would with a find If X is found, do nothing (or update something) Otherwise, insert X at the last spot on the path traversed

Time complexity = O(height of the tree)

delete

When we delete a node, we need to consider how we take care of the children of the deleted node. This has to be done such that the property of the search

tree is maintained.

Delete cont…

Three cases:

(1) the node is a leaf Delete it immediately

(2) the node has one child Adjust a pointer from the parent to bypass that node

Delete cont…

(3) the node has 2 children replace the key of that node with the minimum element at the right

subtree delete the minimum element

Has either no child or only right child because if it has a left child, that left child would be smaller and would have been chosen. So invoke case 1 or 2.

Time complexity = O(height of the tree)

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