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.