Starting Out with Java: From Control Structures through Data Structures (4th Edition) (What's New in Computer Science)
4th Edition
ISBN: 9780134787961
Author: Tony Gaddis, Godfrey Muganda
Publisher: PEARSON
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Chapter 21, Problem 18TF
Program Description Answer
An ancestor of node can never be in a subtree of the node because the root node or starting node is an ancestor of all other nodes in the tree.
Hence, the given statement is “True”.
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2. Write the code for preorder traversal of a binary tree.
struct node{
int data;
struct node *left, *right;
}
void preorder(node *x) {
}
C++
#ifndef BT_NODE_H#define BT_NODE_H
struct btNode{ int data; btNode* left; btNode* right;};
// pre: bst_root is root pointer of a binary search tree (may be 0 for// empty tree) and portArray has the base address of an array large// enough to hold all the data items in the binary search tree// post: The binary search tree has been traversed in-order and the data// values are written (as they are encountered) to portArray in// increasing positional order starting from the first elementvoid portToArrayInOrder(btNode* bst_root, int* portArray);void portToArrayInOrderAux(btNode* bst_root, int* portArray, int& portIndex);
// pre: (none)// post: dynamic memory of all the nodes of the tree rooted at root has been// freed up (returned back to heap/freestore) and the tree is now empty// (root pointer contains the null address)void tree_clear(btNode*& root);
// pre: (none)// post: # of nodes contained in tree rooted at root is returnedint…
Chapter 21 Solutions
Starting Out with Java: From Control Structures through Data Structures (4th Edition) (What's New in Computer Science)
Ch. 21.1 - Prob. 21.2CPCh. 21.1 - Prob. 21.3CPCh. 21 - Prob. 1MCCh. 21 - Prob. 2MCCh. 21 - Prob. 3MCCh. 21 - Prob. 4MCCh. 21 - Prob. 5MCCh. 21 - Prob. 6MCCh. 21 - Prob. 7MCCh. 21 - Prob. 8MC
Ch. 21 - Prob. 9MCCh. 21 - Prob. 10MCCh. 21 - Prob. 11TFCh. 21 - Prob. 12TFCh. 21 - Prob. 13TFCh. 21 - Prob. 14TFCh. 21 - Prob. 15TFCh. 21 - Prob. 16TFCh. 21 - Prob. 17TFCh. 21 - Prob. 18TFCh. 21 - Prob. 19TFCh. 21 - Prob. 20TFCh. 21 - Prob. 21TFCh. 21 - Prob. 1FTECh. 21 - Prob. 2FTECh. 21 - Prob. 3FTECh. 21 - Prob. 1SACh. 21 - Prob. 2SACh. 21 - Prob. 3SACh. 21 - Prob. 4SACh. 21 - What is a priority queue?Ch. 21 - Prob. 6SACh. 21 - Prob. 7SACh. 21 - Prob. 1AWCh. 21 - Prob. 2AWCh. 21 - Prob. 3AWCh. 21 - Prob. 4AWCh. 21 - Prob. 5AWCh. 21 - Prob. 6AWCh. 21 - Prob. 7AWCh. 21 - Prob. 4PCCh. 21 - Prob. 6PC
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- A tree is implemented using a node structure defined as: struct node{ int data; struct node *left; strict node *right; }; Write a function whose prototype is: int smallest(struct node *); which accepts a tree (pointer to the root) and returns the smallest node in the right sub-tree of the root. If root is NULL, or if there is no right sub-tree, returm the number -999.arrow_forwardAssume the tree node structure is following........ struct node { int data; struct node* left; struct node* right; }; struct node *root = null; and there is a created new node function, called newnode(int new_data). Please filled the Blank of Insertion function. void insert(struct node *root, int key) { struct node *current; queue q; q.enque(root); while(!q.empty() } current = q.front(); q.deque(); if(current->left == NULL) { break; } else } q.enque( if(current->right == NULL) { break; else q.enque(_ = newnode(key); = newnode(key); _-));arrow_forward4. Complete the fuction definition given below that takes the root node of a tree as a parameter and returns the count of nodes. int count_nodes(Node *root){ // complete the function }arrow_forward
- True or False: The largest element in any non-empty BST always has no right child. True or False: The smallest element in any non-empty BST always has no right child. A tree node that has no children is called a node. The children of a same parent node are called asarrow_forwardTrue or false: A full tree expressed as an array needs around three times less space than a linked node representation of the same tree?arrow_forwardBreadth-first search must be done without recursion. (with iterative)arrow_forward
- Programming questions：typedef struct node { int data; struct node *left, *right;}BT;The node structure of the binary tree (BT) is shown above. There is a binary tree T, please complete the function: int degreeone(BT *T) to compute how many degree 1 node in the BT. The T is the root pointer, and the function shoule return the total number of degree 1 node.arrow_forwardGiven the following struct that represents a binary tree: struct Node { int key: Node "parent; Node "left; Node "right; Nodelint k) : key(k), parent(nullptr), left(nullptr), right(nullptr) (I: 1: Write a recursive function that prints out the nodes in a tree stored using the above structure in order to cout. The function prints the depth (root depth is at 0) and key of that node separated by a colon (Example "O: 10\n" for root with key 10). Your function CAN NOT create any local variables and can only use what is passed to the function. Use the below function signature (NOTE: this is not a class method). void inorderAndDepth(Node "node, int depth)arrow_forwardstruct insert_into_bst { // Function takes a constant Book as a parameter, inserts that book indexed by // the book's ISBN into a binary search tree, and returns nothing. void operator()(const Book& book) { // // TO-DO (7) ||| ///// // Write the lines of code to insert the key (book's ISBN) and value // ("book") pair into "my_bst". END-TO-DO (7) | } std::map& my_bst; };arrow_forward
- Programming language C++ problemarrow_forwardBinary Search Tree Empirical and Theoretical ResultsPart 1: we need to define a binary search tree data structure. Also, we need to implement the following functions:1. Insert Sorted: BSTREE insert(BSTREE root, int num): root points to a node in a binary search tree; num is a number to be inserted in the tree rooted at “root”. This function returns the root of the modified tree.2. Print Elements: void inorder traversal(BSTREE root, FILE *fp): root points to a node in a binary search tree. This function does not return anything, but prints out, to the file specified, the nodes in the tree rooted at “root” by performing an inorder traversal. Part 2: Test the performance of the designed data structure using theoretical and experimental approaches as follows:1. Dataset 1-Dataset is sorted- Add code to insert the numbers 1...n in that order in an initially empty doubly linked list and a binary search tree.a. Run it on different values of n where :i. n = 20,000ii. n = 50,000iii. n =…arrow_forwardWrite the following functions: a) The function gets a root of a Binary Tree of ints, and a boolean predicate pred. It returns a pointer to a vertex v for which the data satisfies pred, i.e. pred(v->>data)==true. if no such vertex is not found, the function returns -1. If there are several such vertices, the function returns the first such v in pre-order . // finds the first node in pre-order satisfying pred // if such node not found, returns NULL BTnode_t* find(const BTnode_t* root, bool (*pred)(int)); struct BTnode { int value; struct BTnode*left; struct BTnode*right; struct BTnode*parent; }; typedef struct BTnode BTnode_t; BTNODE.C: BTnode_t* create_node(int val) { BTnode_t* newNode = (BTnode_t*) malloc(sizeof(BTnode_t)); newNode->value = val; newNode->left = NULL; newNode->right = NULL; newNode->parent = NULL; returnnewNode; } void set_left_child(BTnode_t* parent, BTnode_t* left_child) { if (parent) parent->left = left_child; if (left_child)…arrow_forward
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