This is what I have to do
We will initialize the tree by hard coding the values illustrated below. You will need to write, test and execute a driver program that will use the complete binary tree implementation.
// this will initialize the tree node values
treenodes[0]=5;
treenodes[1]=10;
treenodes[2]=2;
treenodes[3]=5;
treenodes[4]=7;
treenodes[5]=12;
treenodes[6]=3;
treenodes[7]=9;
treenodes[8]=8;
treesize=9;
print leftchild(3);
print height();
print isleaf(7);
print isleaf(2);
print parent(5);
traverseinorder();
This is what I got:
// CPP btree.cpp : Defines the entry point for the console application. // #include "stdafx.h" class BTree { public: int treenodes[1000]; // structure to store the tree nodes values int size; // the number of nodes in the tree BTree(void); // the "constructor" bool isleaf(int nodeindex); // returns true if the node is a leaf int parent(int nodeindex); // returns the index of the parent int leftchild(int nodeindex); // returns the index of the left child int rightchild(int nodeindex); // returns the index of the right child int height(); // returns the height of the tree int traverseinorder(int); // a recursive function to print the nodes values based on an inorder traversal }; bool BTree::isleaf(int nodeindex) //returns true if the node is a leaf { return (BTree::leftchild(nodeindex)==NULL && BTree::rightchild(nodeindex)==NULL) ; } int BTree::parent(int nodeindex) // returns the index of the parent { int rv =0; rv = (nodeindex - 1) / 2; if (rv >= size) rv = -1 ; return rv ; } int BTree::leftchild(int nodeindex) // returns the index of the left child { int rv =0; rv = (nodeindex +1) *2 - 1; if (rv >= size) rv = -1 ; return rv ; } int BTree::rightchild(int nodeindex) // returns the index of the right child { int rv =0; rv = (nodeindex +1) * 2; if (rv <= size) rv = -1 ; return rv ; } int BTree::traverseinorder(int nodeindex) // a recursive function to print the nodes values based on an inorder traversal { if (nodeindex == -1) return -1 ; return traverseinorder(leftchild(nodeindex)) ; //print the treenode values cout << "Tree node " << nodeindex << " = " << treenodes[nodeindex] << endl; traverseinorder(rightchild(nodeindex)); } // end traverseinorder() int BTree::height() { return 0; } int main(int argc, char* argv[]) { BTree tree1; tree1.treenodes[0]=5; tree1.treenodes[1]=10; tree1.treenodes[2]=2; tree1.treenodes[3]=5; tree1.treenodes[4]=7; tree1.treenodes[5]=12; tree1.treenodes[6]=3; tree1.treenodes[7]=9; tree1.treenodes[8]=8; //set the size here. size is a member of the bTree class so we need our object.member syntax tree1.size = 9; cout << "The left child of node 3 is: " << tree1.leftchild(3) << endl; //cout << "The height of the tree is: " << tree1.height() << endl; cout << "The node 7 is a leaf: " << tree1.isleaf(7) << endl; cout << "The node 2 is a leaf: " << tree1.isleaf(2) << endl; cout << "The parent of node 5 is: " << tree1.parent(5) << endl; cout << "Values for inorder traversal: " << endl; cout << tree1.traverseinorder(0); return 0; }If I put in the Btree(void); it get CPP btree error LNK2019: unresolved external symbol "public: __thiscall BTree::BTree(void)" (??0BTree@@QAE@XZ) referenced in function _main
Ok so I take it out. With it out I have a few problems with the inordertraversal. One if I take out the search of the rightchild it works. If I put in the rightchild it gives me really wierd values. I even tried to take out the left and all I get on the right in 0 = 5? This tells me I'm getting a rightside tree. I still don't have the height done either.
