7. Trees - SKKUmodsim.skku.ac.kr/bbs/Course/Data/DS/LectureNote/07.Tree.pdf · 2018. 4. 10. · Types of Binary Trees Full Binary Tree Degree of every node except leaves is ‘2’

7. Trees

Human’s View

root

leaf

Computer’s View

root

leaf

Tree : Introduction

Array, Stack, Queue, Link list come under Linear Structure

Tree is Non-linear Structure

Very Efficient when using it for Search, Insert, Delete, etc.

Data are organized in a Hierarchical manner.

One common example is in the investigation of genealogies

The pedigree chart shows someone’s ancestors

The lineal chart is a chart of descendants rather than ancestors

Dusty

Honey Bear Brandy

Brunhilde Terry Coyote Nugget

Italic

Osco-Umbrain Latin

Oscan Umbrain Spanish French Italian

Dusty’s father

Dusty’s mother

What is a Tree?

Definition

A tree is a finite nonempty set of nodes such that

1) there is a special node called root

2) remaining nodes are partitioned into n ( 0) disjoint sets

T1, T2, · · ·,Tn where each of these is also a tree;

(Especially, we call each Ti a subtree of the root)

Properties

All nodes must be connected.

At least, one node must exist

Any cycle should not be allowed.

It must have a hierarchical structure

A

B C

D E F G H

Tree : Example

Node A is a root; it has 3 subtrees T1, T2, T3

where T1 = {B, E, F, K, L}, T2 = {C, G}, T3 = {D, H, I, J, M}.

Node B is a root of T1; it has two subtrees T11 and T12

where T11 = {E, K, L}, T12 = {F}.

The remaining nodes are similarly defined.

A

B C

E F

K

G

L

D

H I

M

J

Tree : Terminologies

Degree of a node : The number of subtrees of the node

Degree of a tree : The maximum degree in the tree

Parent / Child : One to Many (1 : m) relationship

Sibling : Child nodes with the same parent

Leaf : node with degree 0

Ancestor : All the nodes along the path from the root to the node

Descendant : All the nodes that are in its subtrees

Level : Level of root is 1;

For subsequent nodes, the level is the node’s parent + 1

Height (= Depth) of a tree : the maximum level of the tree

Tree : Terminologies

Consider a node E; What is its parent? child? sibling? ancestor?

What are leaf nodes? Non-leaf nodes?

What is degree of tree?

What is height of tree?

A

B C

E F

K

G

L

D

H I

M

J

3 Degree

2

2 0

0 0

1

0

3

1 0 0

0

Level

1

2

3

4

What is a Binary Tree?

Definition

A binary tree is a finite set of nodes such that

it is either empty or it consists of root and two disjoint binary trees, called, left subtree and right subtree

Key Characteristics

Degree of any given node must not exceed two. (i.e., 0, 1, or 2)

Left subtree and right subtree are distinguished.

The order/sequence of subtrees is important.

Binary tree may be empty (i.e., zero node)

≠

A

B Two different binary trees, but viewed as trees, they are the same

A

B

B

D E

C

F

A

Left subtree Right subtree

Binary Tree : Examples

(a) (b)

A

(c) (d)

A

B

A

B

(e)

B

D E

C

A

(f)

B

C

A

(g)

B

C

A

Binary Tree : Properties (1)

Number of Nodes of a Binary Tree

Maximum Number of Nodes on level i : 2i–1 (for i ≥ 1)

Maximum Number of Nodes of height h : 2h – 1 (for h ≥ 1)

Minimum Number of Nodes of height h : h (for h ≥ 1)

level

1

2

3

…

h. . . . . . . . . . . . .

Max. Number of Nodes

= 1 + 2 + 4 + … + 2h-1 = 2h–1

hei

ght =

h

level

1

2

3

…

h. . .

Min. Number of Nodes

= 1 + 1 + … + 1 = h

Height (= h) of a Binary Tree

Let n be the number of nodes of a binary tree; Then, h n 2h – 1

1) From h n, maximum height = n

2) From n 2h – 1, minimum height = log2(n+1)

Thus, log2(n+1) h n


level

1

2

3

…

h

. . . h ≤ n

. . . . . . . . . . . . .

. . . . .

level

1

2

3

…

h

n ≤ 2h–1

Relationship between n0 and n2

Let n0 : number of leaf nodes

n1 : number of nodes with degree 1

n2 : number of nodes with degree 2

Then, n0 = n2 + 1

Proof) n : total number of nodes;

Clearly, n = n0 + n1 + n2 ……(1)

Let B : number of branches;

Then, n = B + 1 where B = n1 + 2n2;

Thus, n = n1 + 2n2 + 1 ………(2)

From (1), (2), n0 = n2 + 1


Ex:

n0 = 4, n1 = 2, n2 = 3

B = n1 + 2n2 = 8

Types of Binary Trees

Full Binary Tree

Degree of every node except leaves is ‘2’

All leaves are located at the same level h

Total number of nodes is 2h – 1

Complete Binary Tree

It is a full binary tree down to the level (h – 1)

At the level h, nodes are filled from left to right sequentially.

Skewed(= Degenerated) Binary Tree

Degree of every node except leaves is ‘1’

The tree is skewed or zigzagged

Total number of nodes is h (h is height)

Full Binary Tree

Skewed(= Degenerated) Binary Tree

Binary Trees : Examples

(a) height = 2 (b) height = 3 (c) height = 4

???

A

B

C

D

A

B C

A

B

D E

C

F G

A

B

C

D

(a) (b)

Complete Binary Tree

Not Complete Binary Tree

Binary Trees : Examples

(a) (b) (c)A

B

A

B

D E

C

F

(a) (b)

A

B

D E

C

F G

H I

A

B

D E

C

F G

H I J

A

B

D E

C

F

G H

Binary Tree : Numbering Nodes

Numbering scheme in a Binary Tree

Numbering nodes from top to bottom (in a tree)

Number nodes from left to right (at the same level)

Cf) The numbers are used for recognizing each position at the tree

1

2 3

4 5 6 7

8 9 10 11 12 13 14 15

Position of Parent and Children

Parent of node i is the node floor(i/2), where i ≠ 1

Left child of node i is the node 2i, where 2i n

Right child of node i is the node 2i+1, where 2i + 1 n)

Binary Tree: Parent & Children Position

1

2 3

4 5 6 7

8 9 10 11 12 13 14 15

Ex; for node 5,- Its parent is node 2- Its left child is node 10- Its right child is node 11

Binary Tree Implementation: Array

How to Implement Binary Tree?

Use of One-dimensional Array such that

its size becomes at most 2h if Binary tree has h height (and its 0th

position is not used)

Store elements sequentially from the Root to the Leaves.

Each position of Parent, Left child, Right child of node i is given as

PARENT(i) = floor(i/2) (where, i ≠ 1)

LEFT_CHILD(i) = 2i (where, 2i n)

RIGHT_CHILD(i) = 2i + 1 (where, 2i + 1 n)

What are the Parent, Left child, Right child of node e in tree[ ]? Which positions are they located in?


1

2 3

4 5 6 7

8 9 10

a

b c

d e f g

h i j

a b c d e f g h i jtree[ ]

1 2 3 4 5 6 7 8 9 10

Any Problem?

For Some Cases(e.g., Skewed Binary Tree), many unused spaces exist!


1

3

7

a

b

c

d

15

a - b - - - c - - -tree[ ]

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

- - - - d

In case of n nodes, the array size of binary tree has

Minimum: n + 1 (for full/complete tree)

Maximum: 2n (for skewed tree)

In case of h height, the array size of binary tree has

Maximum: 2h – 1 (excluding 0th position)

If Binary tree is not full or complete, many empty spaces may be incurred within the array.

Especially, a binary tree is skewed, memory(space) utilization is n/2n

Ex) if n is 10, what is the utilization?


. . . . . . . .

. . .

n nodes

h =

nh =

log

(n+

1)

Node Definition

typedef struct node *tree_ptr;

typedef struct node {

int data;

tree_ptr left_child;

tree_ptr right_child;

}

Non-Leaf Node Leaf Node

☞ Links of all leaves are NULL!

Binary Tree Implementation: Linked List

DATAleft_child

right_child

Indicate Left child

Indicate Right child

DATANULL NULL

Example 1


Binary Tree Linked List Representation

A

B

D E

C

F G

H I

A

B

root

C

D E

NULL

NULL F

NULL

NULL G

NULL

NULL

HNULL

NULL I

NULL

NULL


Example 2

Binary Tree Linked List Representation

A

B

C

D

E

root

D

NULL

E

NULL

NULL

C

NULL

B

NULL

A

NULL

Use of Array

(Different from Binary tree), the use of arrayis hard to find the positions corresponding toparent and children of the given node

Use of Linked List (more general)

Number of link fields is identical to the maximum degree k of the treesince each node’s degree is usually different

Every nodes whose degree is less than k have NULL links

Leaf has all NULL links

Tree Implementation

child 1 child 2

DATA

… …

child k… …

a b - - c - -

tree[ ] 1 2 3 4 5 6 7

Too Many Null Links in a Tree

We need k links per each node (where k : degree of a tree)

Number of links : n * k (where n : number of nodes in a tree)

Number of non-null links : n – 1

There exist one link that points to each node (excluding the root)

Number of null links : (n * k) – (n – 1)

Null links percentage : ((k – 1) * n + 1) / (k * n)

Ex1) if k = 10, then approx. 90%

Ex2) if k = 2, then approx. 50%

Tree Implementation

A

B C

D E F G H

For instance, n=8, k=3

(k, #non-null)(3,2)

(3,2) (3,3)

(3,0) (3,0) (3,0) (3,0) (3,0)

Binary Tree Representation to a Tree

It consists of Two Procedures: Convert & Rotate

(1) Convert each node in a tree by choosing its left-most child and its right-next-sibling;

(2) Rotate the left-child/right-sibling tree by 45 degrees clockwise;

Note : - Right link of the root node always has NULL value.

- NULL link percentage considerably decreases.

child 1 child 2

DATA

… …

child k… … Left-most-child Right-next-Sibling

DATA

Tree

Left child-Right sibling

Tree

Binary Tree

convert


A

B

E G

D

F

C

A

B

E G

D

F

C

A

B

E

G

DF

C


Tree Left child-Right sibling Tree Binary Tree

A

B

A

B

A

B

A

B C

Tree Left child-Right sibling Tree Binary Tree

A

B C

A

B

C

convert rotate

convert rotate

Binary Tree Traversal


It comes under visiting each node in a binary tree exactly once.

It produces a linear order for the information in a binary tree.

Traversal methods

INORDER : Left, Visit, Right (LVR)

PREORDER : Visit, Left, Right (VLR)

POSTORDER : Left, Right, Visit (LRV)

LEVEL-ORDER

(Where, Left moves to the left of the current node

Visit visits to the current node

Right moves to the right of the current node)

Procedures

Inorder : LVR

ExampleINORDER : B A C

PREORDER: A B C

POSTORDER: B C A

A

B C

Preorder : VLR

2

1 3

left subtree right subtree

1

2 3


3

1 2


Postorder : LRV


INORDER : Recursive Version

INORDER (tree_ptr ptr) {

if (ptr != NULL)

{

INORDER (ptr -> left_child)

printf (ptr -> data)

INORDER (ptr -> right_child)

}

}

INORDER Traversal in a Binary Tree

ptr points to the first node in a tree

Inorder: LVR

a

b c

d e f

Inorder : LVR

4

2

6

1 3 5

Result: d b e a f c

*

+

E

* D

/ C

A B

6 7 9 10

12 13

15 16

18 19

5

4

3

2

1

17

14

11

8

Output: A / B * C * D + E

INORDER (LVR) : ExampleTrace of ProgramBinary Tree with Arithmetic Expression

Inorder (Value)

ValueInorder (Value)

Value

1 (+) - 11 (C) -

2 (*) - 12 (Null) -

3 (*) - 11 (C) printf

4 (/) - 13 (Null) -

5 (A) - 2 (*) printf

6 (Null) - 14 (D) -

5 (A) Printf 15 (Null) -

7 (Null) - 14 (D) printf

4 (/) Printf 16 (Null) -

8 (B) - 1 (+) printf

9 (Null) - 17 (E) -

8 (B) Printf 18 (Null) -

10 (Null) - 17 (E) printf

3 (*) Printf 19 (Null) -

INORDER (tree_ptr ptr) {

tree_ptr stack [MAX_STACK_SIZE];

top = -1; /* Initialize stack */

for( ; ; ) {

while (ptr != NULL) {

top = top + 1;

if (top >= MAX_STACK_SIZE–1) return;

stack[top] = ptr;

ptr = ptr -> left_child; }

if (top == -1) break; /*empty stack*/

ptr = stack[top];

top = top – 1;

printf(ptr -> data);

ptr = ptr -> right_child;

}

}

INORDER (LVR) : Iterative Version

*

/ C

A B

4 5 7 8

10 113

2

1

9

6

A

/

stack

pop

*

pop B

stack

*

poppop

stack

C pop

Result:A /

Result:A / B *

Result:A/B*C

INORDER : Performance Analysis

For the Iterative Version

We need a stack to return to previously visited nodes safely.

Every node in a binary tree is placed on and removed from the stack once.

Time Complexity

Let n be the number of nodes in a binary tree

Every node is pushed and popped exactly once; i.e., n times

Also, ptr will be NULL once for every (n + 1) NULL links

Thus, the time complexity is O(n)

Space Complexity

The stack size required is equal to the height h of the binary tree;

log2(n+1) h n

Thus, the space complexity becomes O(n)

PREORDER

PREORDER (tree_ptr ptr) {

if (ptr != NULL)

{

printf(ptr->data)

PREORDER(ptr->left_child)

PREORDER(ptr ->right_child)

}

}

PREORDER Traversal in a Binary Tree


Preorder: VLR

a

b c

d e f

Preorder : VLR

1

2 5

3 4 6

Result: a b d e c f

POSTORDER (tree_ptr ptr) {

if (ptr != NULL)

{

POSTORDER(ptr->left_child)

POSTORDER(ptr ->right_child)

printf(ptr->data)

}

}

POSTORDER

POSTORDER Traversal in a Binary Tree


Postorder: VLR

a

b c

d e f

Postorder : LRV

6

3

5

1 2 4

Result: d e b f c a

LEVEL-ORDER (= Breadth First)

LEVEL_ORDER (tree_ptr ptr) {

int front = rear = 0;

tree_ptr queue[MAX_QUE_SIZE];

if (ptr) insert(ptr); /*Insert ptr into queue*/

while (ptr != null) {

ptr = delete(); /*Delete from the queue */

printf(ptr -> data);

if (ptr -> left_child )

insert(ptr -> left_child);

if (ptr -> right_child )

insert(ptr -> right_child);

}

}

LEVEL-ORDER Traversal

It visits the nodes using the orders by the numbering scheme (for trees):

It traverses a tree top to bottom and left to right.

a

b c

d e f

Lev

el-O

rder

3

4 5 6

Result: a b c d e f

1

2

Queue

abc

ptr

d

Insert delete

e

PREORDER (VLR) : a b d g h e i c f j

POSTORDER (LRV) : g h d i e b j f c a

LEVEL-ORDER : a b c d e f g h i j

INORDER (LVR) : g d h b e i a f j c

Example

a

b c

d e f

jg h i

INORDER : (a + b) * (c – d) / (e + f)

PREORDER : / * + a b – c d + e f

POSTORDER : a b + c d – * e f + /

/

* +

+ - e f

a b c d

Expression Tree

Infix form!

Prefix form!

Postfix form!

Binary Tree Reconstruction

Possible to Reconstruct a Binary Tree from its Traversal results?

☞ Yes! But we need two traversal results.

Available Combinations for Reconstruction

INORDER & PREORDER

INORDER & POSTORDER

INORDER & LEVEL-ORDER

Root Position in Traversal Results

PREORDER & LEVEL-ORDER : the Root exists at the first position

POSTORDER : the Root exists at the last position

INORDER : the Root exists somewhere in the result

PREORDER : a b d g h e i c f j

POSTORDER : g h d i e b j f c a

LEVEL-ORDER : a b c d e f g h i j

INORDER : g d h b e i a f j c

a

b c

d e f

jg h i

Root of T

Subtree:T1

Subtree:T2

Binary Tree Reconstruction

Tree: T

T1 T2

Root of T1

Root of T2

With INORDER & PREORDER

What is the Binary Tree for the Traversal results of INORDER & PREORDER?



Scan PREORDER left to right using INORDER to separate left and right subtrees.

a is the root of the tree; {g, d, h, b, e, i} are in the left subtree; {f, j, c} are in the right subtree.

{ g, d, h, b, e, i } { f, j, c }

a



b is the next root (of the left branch); {g, d, h} are in the left subtree; {e, i} are in the right subtree.

{ g, d, h, b, e, i } { f, j, c }

a

a

b { f, j, c }

{ g, d, h } { e, i }



d is the next root; g is the left subtree; h is the right subtree.

a

b { f, j, c }

{ g, d, h } { e, i }

a

b

d

{ f, j, c }

{ e, i }

{ g } { h }


{g, h} has no child


e is the next root; i is the right subtree


c is the next root; {f, j} is in the left subtree;

f is the next root; j is in the right subtree.

a

b

d

{ f, j, c }

{ e, i }

{ g } { h }

e

{ i }

c

f

{ j }

With INORDER & POSTORDER

What is the Binary Tree for the Traversal results of INORDER & POSTORDER?



Scan POSTORDER right to left using INORDER to separate left and right subtrees.

a is the root of the tree;{g, d, h, b, e, i} are in the left subtree; {f, j, c} are in the right subtree.

{ g, d, h, b, e, i } { f, j, c }

a

With INORDER & POSTORDER


c is the next root (of the right branch);There is no right subtree; {f, j} are in the left subtree.

{ g, d, h, b, e, i } { f, j, c }

a

a

c{ g, d, h, b, e, i }

{ f, j }

Do the remaining

steps!

Binary Tree Traversal : Copy

tree_ptr COPY (tree_ptr original) {

tree_ptr temp;

if (original != NULL) {

temp = (tree_ptr) malloc (sizeof(node));

if ( IS_FULL(temp) ) exit;

temp -> left_child = COPY(original -> left_child)

temp -> right_child = COPY(original -> right_child);

temp -> data = original -> data;

return temp;

}

return NULL;

}

Copy a Binary Tree by using POSTORDER

Null Null

d

temp->left_child->left_child=COPY( )

temp->left_child=COPY( )

Null Null

b

Null Null

e

a

b c

d e f

Null Null Null Null Null Null

temp->left_child->left_child->left_child=Null

…

Documents

7. Trees - SKKUmodsim.skku.ac.kr/bbs/Course/Data/DS/LectureNote/07.Tree.pdf · 2018. 4. 10. · Types of Binary Trees Full Binary Tree Degree of every node except leaves is ‘2’