1 Chapter 10 Trees. 2 Definition of Tree A tree is a set of linked nodes, such that there is one and...

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Chapter 10Trees

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Definition of Tree

• A tree is a set of linked nodes, such that there is one and only one path from a unique node (called the root node) to every other node in the tree.

• A path exists from node A to node B if one can follow a chain of pointers to travel from node A to node B.

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A

E

F

B

C

D

G

A set of linked nodes

Paths

There is one path from A to B

There is a path from D to B

There is also a second path from D to B.

4

A

E

F

B

C

D

G

There is no path from C to any other node.

Paths (cont.)

5

Cycles

• There is no cycle (circle of pointers) in a tree.

• Any linked structure that has a cycle would have more than one path from the root node to another node.

6

Example of a Cycle

A

C

D

B

E

C → D → B → E → C

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Tree Cannot Have a Cycle

A

C

D

B

E

2 paths exist from A to C:1. A → C2. A → C → D → B → E → C

8

Example of a Tree

A

C B D

E F G

root

In a tree, every pair of linked nodes have a parent-child relationship (the parent is closer to the root)

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Example of a Tree(cont.)

A

C B D

E F G

root For example, C is a parent of G

10


A

C B D

E F G

root E and F are children of D

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A

C B D

E F G

root The root node is the only node that has no parent.

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A

C B D

E F G

rootLeaf nodes (or leaves for short) have no children.

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Subtrees

A

B C

I K D E F

J G H

subtree

root A subtree is a part of a tree that is a tree in itself

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Binary Trees

• A binary tree is a tree in which each node can only have up to two children…

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NOT a Binary Tree

A

B C

I K D E F

J G H

root C has 3 child nodes.

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Example of a Binary Tree

A

B C

I K E

J G H

root The links in a tree are often called edges

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Levels

A

B C

I K E

J G H

rootlevel 0

level 1

level 2

level 3

The level of a node is the number of edges in the path from the root node to this node

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Full Binary Tree

B

D

H

root

I

A

E

J K

C

F

L M

G

N O

In a full binary tree, each node has two children except for the nodes on the last level, which are leaf nodes

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Complete Binary Trees

• A complete binary tree is a binary tree that is either– a full binary tree– OR– a tree that would be a full binary tree but it is

missing the rightmost nodes on the last level

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NOT a Complete Binary Trees

B

D

H

root A

E

C

F G

I Missing non-rightmost nodes on the last level

21

Complete Binary Trees(cont.)

B

D

H

root

I

A

E

J K

C

F

L

G

Missing rightmost nodes on the last level

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Complete Binary Trees(cont.)

B

D

H

root

I

A

E

J K

C

F

L M

G

N O

A full binary tree is also a complete binary tree.

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Binary Search Trees

• A binary search tree is a binary tree that allows us to search for values that can be anywhere in the tree.

• Usually, we search for a certain key value, and once we find the node that contains it, we retrieve the rest of the info at that node.

• Therefore, we assume that all values searched for in a binary search tree are distinct.

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Properties of Binary Search Trees

• A binary search tree does not have to be a complete binary tree.

• For any particular node, – the key in its left child (if any) is less than its

key.– the key in its right child (if any) is greater than

or equal to its key.• Left < Parent <= Right.

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Binary Search Tree (BST)Node

template <typename T>BSTNode {

T info;BSTNode<T> *left;BSTNode<T> *right;

};

The implementation of a binary search tree usually just maintains a single pointer in the private section called root, to point to the root node.

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Inserting Nodes Into a BST

37, 2, 45, 48, 41, 29, 20, 30, 49, 7

Objects that need to be inserted (only key values are shown):

root: NULL

BST starts off empty

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Inserting Nodes Into a BST (cont.)

37, 2, 45, 48, 41, 29, 20, 30, 49, 7

root 37

28


2, 45, 48, 41, 29, 20, 30, 49, 7

root 37

2 < 37, so insert 2 on the left side of 37

29


2, 45, 48, 41, 29, 20, 30, 49, 7

root 37

2

30


45, 48, 41, 29, 20, 30, 49, 7

root 37

2

45 > 37, so insert it at the right of 37

31


45, 48, 41, 29, 20, 30, 49, 7

root 37

2 45

32


48, 41, 29, 20, 30, 49, 7

root 37

2 45

When comparing, we always start at the root node

48 > 37, so look to the right

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48, 41, 29, 20, 30, 49, 7

root 37

2 45

This time, there is a node already to the right of the root node. We then compare 48 to this node

48 > 45, and 45 has no right child, so we insert 48 on the right of 45

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48, 41, 29, 20, 30, 49, 7

root 37

2 45

48

35


41, 29, 20, 30, 49, 7

root 37

2 45

4841 > 37, so look to the right

41 < 45, so look to the left – there is no left child, so insert

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41, 29, 20, 30, 49, 7

root 37

2 45

4841

37


29, 20, 30, 49, 7

root 37

2 45

484129 < 37, left

29 > 2, right

38


29, 20, 30, 49, 7

root 37

2 45

484129

39


20, 30, 49, 7

root 37

2 45

48412920 < 37, left

20 > 2, right

20 < 29, left

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20, 30, 49, 7

root 37

2 45

484129

20

41


30, 49, 7

root 37

2 45

484129

2030 < 37

30 > 2

30 > 29

42


30, 49, 7

root 37

2 45

484129

20 30

43


49, 7

root 37

2 45

484129

20 30

49 > 3749 > 4549 > 48

44


49, 7

root 37

2 45

484129

20 30 49

45


7

root 37

2 45

484129

20 30 49

7 < 377 > 27 < 297 < 20

46


7

root 37

2 45

484129

20 30 49

7

47


root 37

2 45

484129

20 30 49All elements have been inserted

7

48

Searching for a Key in a BST

root 37

2 45

484129

20 30 49

Searching for a key in a BST uses the same logic

7Key to search for: 29

49

Searching for a Key in a BST (cont.)root

37

2 45

484129

20 30 49

Key to search for: 29

29 < 37

7

50


37

2 45

484129

20 30 49


29 > 2

7

51


37

2 45

484129

20 30 49


29 == 29

FOUND IT!

7

52


37

2 45

484129

20 30 49

Key to search for: 3 7

53


37

2 45

484129

20 30 49


3 < 7

7

3 < 20

3 < 29

3 > 2

3 < 37

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37

2 45

484129

20 30 49


When the child pointer you want to follow is set to NULL, the key you are looking for is not in the BST

7

55

Searching a BST is Quicker than Array or Linked List

root 37

2 45

484129

20 30 49

Search for a key in this BST takes max 5 comparisons (key = 7)

7

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Searching a BST is Quicker than Array or Linked List (cont.)

37, 2, 45, 48, 41, 29, 20, 30, 49, 7

Searching for a key in linear data structures, such as array or linked list, takes max 10 comparisons (key = 7). In fact searching for key 29, 20, 30, 49, and 7 all takes more than 5 comparisons. Overall slower than BST.

The main advantage of BST over linear data structures is quicker search.

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Deleting a BST Node

• Deleting a node in a BST is a little tricky – it has to be deleted so that the resulting structure is still a BST with each node greater than its left child and less than its right child.

• Deleting a node is handled differently depending on whether the node:– has no children– has one child– has two children

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Deletion Case 1: No Children

root 37

2 45

484129

20 30 49

Node 49 has no children – to delete it, we just remove it

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Deletion Case 1: No Children (cont.)

root 37

2 45

484129

20 30

60

Deletion Case 2: One Child

root 37

2 45

484129

20 30 49

Node 48 has one child – to delete it, we just splice it out

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Deletion Case 2: One Child (cont.)

root 37

2 45

484129

20 30 49

Node 48 has one child – to delete it, we just splice it out

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root 37

2 45

4129

20 30 49

63


root 37

2 45

484129

20 30 49

Another example: node 2 has one child – to delete it we also splice it out

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root 37

2 45

484129

20 30 49

Another example: node 2 has one child – to delete it we also splice it out

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root 37

45

484129

20 30 49

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Deletion Case 3: Two Children

root 37

2 45

484129

20 30 49

Node 37 has two children…

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Deletion Case 3: Two Children (cont.)root

37

2 45

484129

20 30 49

to delete it, first we find the greatest node in its left subtree

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37

2 45

484129

20 30 49

First, we go to the left once, then follow the right pointers as far as we can

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37

2 45

484129

20 30 4930 is the greatest node in the left subtree of node 37

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37

2 45

484129

20 30 49Next, we copy the object at node 30 into node 37

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30

2 45

484129

20 49

Finally, we delete the lower red node using case 1 or case 2 deletion

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30

2 45

484129

20 49Let’s delete node 30 now

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30

2 45

484129

20 49

29 is the greatest node in the left subtree of node 30

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30

2 45

484129

20 49

Copy the object at node 29 into node 30

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29

2 45

484129

20 49

This time, the lower red node has a child – to delete it we use case 2 deletion

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29

2 45

484129

20 49

This time, the lower red node has a child – to delete it we use case 2 deletion

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29

2 45

4841

20 49

78

Searching a BST is Quicker than Array or Linked List

root 37

2 45

484129

20 30 49

Search for a key in this BST takes max 5 comparisons (key = 7)

7

: 37 2 29 20 7 30 45 41 48 49

Traversing a BST• There are 3 ways to traversal a BST (visit

every node in BST):– 1. Preorder (parent → left → right)– 2. Inorder (left → parent → right)– 3. Postorder (left → right → parent)

• Result of traversing BST (at pp. 55):– Preorder: – Inorder:– Postorder:

2 7 20 29 30 37 41 45 48 497 20 30 29 2 41 49 48 45 37

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1 template <typename T>2 class BinarySearchTree {3 BSTNode *root;4 void preOrderInternal (BST<T> *parent) { … }5 void inOrderInternal (BST<T> *parent) { … }6 void postOrderInternal (BST<T> *parent) { … }7 void insertInternal (BST<T> *parent, 8 T& newElement) { … }9 bool searchInternal (BST<T> *parent, T& target,10 T& foundElement) { … }

Binary Search Tree Class Template

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11 public:12 BinarySearchTree() { … }13 ~BinarySearchTree() { … }14 bool isEmpty() { … }15 void preOrderTraversal() { … }16 void inOrderTraversal() { … }17 void postOrderTraversal() { … }18 void insert (T element) { … }19 bool search (T element, T& foundElement) { … }20 void makeEmpty() { … }21 }

Binary Search Tree Class Template (cont.)

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1 BinarySearchTree()2 : root(NULL) { }34 ~BinarySearchTree()5 {6 makeEmpty();7 }8 9 bool isEmpty()10 {11 return root == NULL:12 }

Constructor, Destructor, IsEmpty()

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2 void inOrderTraversal()3 {4 inOrderInternal (root);5 }

The client uses the driver called InOrderTraversal.

Printing ElementsIn Order

The recursive function inOrderInternal is called. inOrderInternal prints all element in BST in sorted order, starting from root node.

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2 void inOrderInternal (BSTNode<T> *parent)3 {4 if (parent != NULL) {5 inOrderInternal (parent->left);6 cout << parent->info << " ";7 inOrderInternal (parent->right);8 }9 }

The base case occurs when parent is NULL – just returns.

Printing ElementsIn Order (cont.)

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There are 2 recursive calls under the if – the first advances the pointer to the left child…

and the second advances the pointer to the right child.


86


Each recursive call approaches the base case…

it goes down one level through the tree, and it get closer to the case where parent->left or parent->right is NULL.


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If the BST contains only one node, the root node is the only node – the left and right pointers of the root node will be set to NULL.


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parent->left is NULL, so NULL is passed into parent of the new inOrderInternal function.


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Since parent is NULL in the new inOrderInternal function, it will return right away.


90


parent->right is NULL – thus, the recursive function call immediately returns as before.


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In fact inOrderInternal works correctly when there is only 0, 1 or many node in the BST.

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inOrderInternal is called once from the driver inOrderTraversal. Then, for each node in the tree, 2 calls to inOrderInternal are made, giving us a total of 2n + 1 calls where n is the number of nodes.

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2 void preOrderInternal (BSTNode<T> *parent)3 {4 if (parent != NULL) {5 cout << parent->info << " ";6 preOrderInternal (parent->left);7 preOrderInternal (parent->right);8 }9 }

Printing ElementsPre Order

preOrderInternal prints the parent first.

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2 void postOrderInternal (BSTNode<T> *parent)3 {4 if (parent != NULL) {5 postOrderInternal (parent->left);6 postOrderInternal (parent->right);7 cout << parent->info << " ";8 }9 }

Printing ElementsPost Order

postOrderInternal prints the parent last.

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1 bool search (T target, T& foundElement) {2 return searchInternal (root, target, foundElement);3 }45 void insert ( T& newElement) {6 insertInternal (root, newElement);7 }

Searching and Insertion in BST

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Searching and Insertion in BST

• We can actually figure out searchInternal and insertInternal using the same recursive logic as traversal in BST.

References

• Childs, J. S. (2008). Trees. C++ Classes and Data Structures. Prentice Hall.

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