Starting Out with Java: From Control Structures through Data Structures (4th Edition) (What's New in Computer Science)
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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#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…
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.
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