Notes on Trees - Introduction to Programming II | CS 112, Study notes of Computer Science

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Introduction to Programming II^ Trees^ Chris BrooksDepartment of Computer ScienceUniversity of San Francisco

Department of Computer Science — University of San Francisco – p. 1/

22-2: Trees^ •^ Previously, we’ve talked about how to store objects in a linearsequence.^ ◦^ Arrays^ ◦^ ArrayLists^ ◦^ LinkedLists^ •^ These are nice when we care about keeping everything inorder.^ •^ Finding particular elements can take a while, though.

Department of Computer Science — University of San Francisco – p. 2/

22-4: Tree Terminology^ •^ Parent / Child^ •^ Leaf node^ •^ Root node^ •^ Edge (between nodes)^ •^ Path^ •^ Ancestor / Descendant^ •^ Depth of a node

n ◦ Length of path from root to^ n • Height of a tree ◦ (Depth of deepest node) + 1

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22-5: Implementing a tree^ •^ A struct representing a tree containing strings:^ typedef^ struct^ treeNode

*treeptr; typedef^ struct^ treeNode

{ char^ *data;treeptr^ left;treeptr^ right; } treeNode; • We need to declare a type called ’treeptr’, because otherwisethe C compiler doesn’t know what a treeNode is until thedefinition is processed.

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22-7: Binary Search Trees^ •^ Binary Trees^ •^ For each node n, (value stored at node n)

^ (value stored in left subtree) • For each node n, (value stored at node n)

<^ (value stored in right subtree)

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22-8: Adding methods to BST void^ insert(treeptr^ root,

char^ {*newdata) treeNode^ *newNode;if^ (strcmp(root->data,

newdata)^ <=^ 0)^ { if (root->left == NULL)^ { newNode = makeNode(newdata);root->left = newNode;return; } else { return insert(root->left,^ newdata); } }^ else^ {^ if^ (root->right^ ==^

NULL)^ { newNode = makeNode(newdata);root->right = newNode;return; } else { return insert(root->right,^ newdata); } } }

Department of Computer Science — University of San Francisco – p. 8/

22-10: Finding an Element in a BST^ •^ First, the Base Case – when is it easy to determine if anelement is stored in a Binary Search Tree?^ ◦^ If the tree is empty, then the element can’t be there^ ◦^ If the element is stored at the root, then the element is there

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22-11: Finding an Element in a BST^ •^ Next, the Recursive Case – how do we make the problemsmaller?

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22-13: Finding an Element in a BST^ •^ Next, the Recursive Case – how do we make the problemsmaller?^ ◦^ Both the left and right subtrees are smaller versions of theproblem. Which one do we use?^ ◦^ If the element we are trying to find is

<^ the element stored at the root, use the left subtree. Otherwise, use the rightsubtree.

Department of Computer Science — University of San Francisco – p. 13/

22-14: Finding an Element in a BST^ •^ Next, the Recursive Case – how do we make the problemsmaller?^ ◦^ Both the left and right subtrees are smaller versions of theproblem. Which one do we use?^ ◦^ If the element we are trying to find is

<^ the element stored at the root, use the left subtree. Otherwise, use the rightsubtree. • How do we use the solution to the subproblem to solve theoriginal problem?

Department of Computer Science — University of San Francisco – p. 14/

22-16: Finding an Element in a BST To find an element^ e

in a Binary Search Tree

T^ :

-^ If^ T^ is empty, then

e^ is not in^ T

-^ If the root of^ T^ contains

e, then^ e^ is in^ T

-^ If^ e <^ the element stored in the root of

T^ :

◦^ Look for^ e^ in the left subtree of

T

Otherwise^ ◦^ Look for^ e^ in the right subtree of

T^ Department of Computer Science — University of San Francisco – p. 16/

22-17: Exercise^ •^ Implement the find(treeNode *root, char *object) method.^ •^ Write a main() that inserts names into the tree and tests find().^ •^ Count how many recursive calls it takes to find something.^ •^ How is this related to the number of elements in the tree?

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22-19: Counting nodes^ •^ So how would we count the number of nodes in a tree?

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22-20: Counting nodes^ •^ If the tree is null, return 0.^ •^ Otherwise, return 1 + the number of nodes in the left subtree +the number of nodes in the right subtree.^ •^ Exercise: add a countNodes(treeNode *root) function to ourtree data structure.

Department of Computer Science — University of San Francisco – p. 20/