Lecture10 trees v3

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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.

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A

E

F

B

C

D

G

There is no path from C to any other node.

Paths (cont.)

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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.

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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

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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

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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

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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

its key.

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Binary Search Tree 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

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

root 37

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

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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

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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

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29, 20, 30, 49, 7

root 37

2 45

484129

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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

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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

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37

2 45

484129

20 30 49


29 == 29

FOUND IT!

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37

2 45

484129

20 30 49

Key to search for: 3 7

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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

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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

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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

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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

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

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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)

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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 const T& newElement) { … }9 bool searchInternal (const 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.


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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.


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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 (const 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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Lecture10 trees v3

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