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schmidty169 2 Light Poster

20 Years Ago

I'm getting error C2109: subscript requires array or pointer type

// CPP btree.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"

class TreeNode 
{
public:
         TreeNode(string str): data(str), left(NULL), right(NULL) {} // Constructor.  Make a node containing str.

         string data;       // The data in this node.
         TreeNode *left;    // Pointer to left subtree.
         TreeNode *right;   // Pointer to right subtree.
};

typedef TreeNode* TreeNodePtr;

void treeInsert(TreeNodePtr& root, string newItem) 
{
          if ( root == NULL ) {
             root = new TreeNode( newItem );
             return;
          }
          else if ( newItem < root->data ) {
             treeInsert( root->left, newItem );
          }
          else 
          {
             treeInsert( root->right, newItem );
          }
}  // end treeInsert()

void traverseinorder(TreeNode *root )  // a recursive function to print the nodes values based on an inorder traversal
    {
        if ( root != NULL ) {  
           traverseinorder( root->left );    
           cout << root->data << " ";     
           traverseinorder( root->right );   
        }
     } // end traverseinorder()

int size( TreeNode *root )  // the number of nodes in the tree
     {
        if ( root == NULL )
           return 0;  // The tree is empty.  It contains no nodes.
        else 
        {
           int count = 1;   // Start by counting the root.
           count += size(root->left);  // Add the number of nodes in the left subtree.
                                         
           count += size(root->right); // Add the number of nodes in the right subtree.
                                        
           return count;  
        }
     } // end size()

bool isleaf(TreeNode *root) //returns true if the node is a leaf
{
  return (root->left==NULL && root->right==NULL);
}
 
int parent(int pos)                                // returns the index of the parent
{
    return (pos - 1) / 2;
}

int leftchild(int pos)  // returns the index of the left child
{
return (pos * 2) + 1;
}

int rightchild(int pos)  // returns the index of the right child
{
return (pos * 2) + 2;
}
 
int height(TreeNode *root) // returns the height of the tree
{
  if (root==NULL) 
  {
    return(0);
  }
  else 
  {
    // compute the depth of each subtree
    int lheight = height(root->left);
    int rheight = height(root->right);

    // use the larger one
    if (lheight > rheight) return(lheight+1);
    else return(rheight+1);
  }
}  

int main(int argc, char* argv[])
{

treeInsert[0]=5;
treeInsert[1]=10;
treeInsert[2]=2;
treeInsert[3]=5;
treeInsert[4]=7;
treeInsert[5]=12;
treeInsert[6]=3;
treeInsert[7]=9;
treeInsert[8]=8;


cout << leftchild(3);
cout << height();
cout << isleaf(7);
cout << isleaf(2);
cout << parent(5);
traverseinorder();
}

c++

4 Contributors
8 Replies
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Latest Post 20 Years Ago Latest Post by schmidty169

Dani AI

Generated 6 Months Ago

The original compile error (C2109) came from treating a function like an array (for example treeInsert[0] = 5;). spotted that early. After that moved toward an array-based representation but mixed pointer-style code and integer/index-style code: that’s the root cause of most remaining errors (comparing indexes to NULL, calling root->leftchild, using tree-node pointers and treenodes[] at the same time).

Pick one model and stick to it. Since the code in main fills treenodes[], the simplest fix is to treat the tree as an implicit binary tree stored in an array (indexes 0..size-1). Important rules:

Left child of index i is 2i+1; right child is 2i+2.
A child index is invalid if it is >= size (use -1 as the “no child” sentinel).
The parent of index 0 does not exist (return -1); for others use (i-1)/2.
Don’t compare integer indexes to NULL; NULL is a pointer concept (0) and leads to confusion.

Here are compact, correct helper routines (place inside BTree as members):

int left_index(int i) const { int c = 2*i + 1; return (i<0 || i>=size || c>=size) ? -1 : c; }
int right_index(int i) const { int c = 2*i + 2; return (i<0 || i>=size || c>=size) ? -1 : c; }
int parent_index(int i) const { if (i<=0 || i>=size) return -1; return (i-1)/2; }
bool is_leaf(int i) const { return left_index(i)==-1 && right_index(i)==-1; }

void traverse_inorder(int i=0) const {
  if (i<0 || i>=size) return;
  int L = left_index(i); if (L!=-1) traverse_inorder(L);
  std::cout << treenodes[i] << ' ';
  int R = right_index(i); if (R!=-1) traverse_inorder(R);
}

int height(int i=0) const {
  if (i<0 || i>=size) return 0;
  int lh = height(left_index(i)), rh = height(right_index(i));
  return 1 + (lh>rh ? lh : rh);
}

Quick debugging tips: print indexes while testing, initialize size before using helpers, and consider std::vector<int> instead of a fixed array. ’s request for variable declarations is spot-on — ensure your class members and function signatures match the representation you choose.

All 8 Replies

GloriousEremite 25 Junior Poster in Training

20 Years Ago

treeInsert is a function, not an array...

WolfPack 491 Posting Virtuoso

20 Years Ago

Where are the variable declarations for treeInsert, leftchild and the such?

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schmidty169 2 Light Poster · Answer 1 · 2006-07-15T06:28:14+00:00

Ok I scraped that and started new. Not sure if it is an improvement or not.

// CPP btree.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"

class BTree 
{
    
    struct tree_node 
    {
        tree_node *left;   // left subtree has smaller elements
        tree_node *right;  // right subtree has larger elements
        int data;
    };
tree_node *root;

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
        void traverseinorder();  // a recursive function to print the nodes values based on an inorder traversal
};

BTree::BTree()
{
    root=new tree_node;
    root->left=NULL;
    root->right=NULL;
}

bool BTree::isleaf(int nodeindex) //returns true if the node is a leaf
{
  return (nodeindex->left==NULL && nodeindex->right==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 * 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 ;
}

void traverseinorder(BTree *root )  // a recursive function to print the nodes values based on an inorder traversal
    {
        if ( root != NULL ) {  
           traverseinorder( root->left );    
           cout << root->item << " ";     
           traverseinorder( root->right );   
        }
     } // end traverseinorder()

int height(BTree *root) // returns the height of the tree
{
  if (root==NULL) 
  {
    return(0);
  }
  else 
  {
    // compute the depth of each subtree
    int lheight = height(root->left);
    int rheight = height(root->right);

    // use the larger one
    if (lheight > rheight) return(lheight+1);
    else return(rheight+1);
  }
}  

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;
        tree1.traverseinorder();

        return 0;
}

schmidty169 2 Light Poster · Answer 2 · 2006-07-15T06:53:49+00:00

in the traverse I found and fixed this

cout << root->data << " ";

schmidty169 2 Light Poster · Answer 3 · 2006-07-15T08:14:18+00:00

closer if I can figure out the traverse and height

// 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
        void traverseinorder();  // 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 * 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 ;
}

void traverseinorder()  // a recursive function to print the nodes values based on an inorder traversal
    {
        if ( size!= NULL ) {  
             
           cout <<   " ";     
             
        }
     } // end traverseinorder()

int height() // returns the height of the tree
{
  if (size==NULL) 
  {
    return(0);
  }
  else 
  {
    // compute the depth of leftchild
    int lheight = height();
    
    return(lheight+1);
  }
}  

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;
        tree1.traverseinorder();

        return 0;
}

schmidty169 2 Light Poster · Answer 4 · 2006-07-15T11:34:34+00:00

Ok this is where i am now

// 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
        void traverseinorder();  // 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 ;
}

void BTree::traverseinorder(BTree *root)  // a recursive function to print the nodes values based on an inorder traversal
    {
        if ( root!= NULL ) 
        {  
           traverseinorder(root->leftchild));   
           cout << root << " ";     
           traverseinorder(root->rightchild); 
        }
     } // end traverseinorder()

int BTree::height(BTree *root) // returns the height of the tree
{
 if (root==NULL) 
 {
   return(0);
  }
 else 
 {
    //compute the depth of leftchild
    int lheight = height(root->leftchild);
    
    return(lheight+1);
  }
}  

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;
    //    tree1.traverseinorder();

        return 0;
}

Eddy Dean 13 Junior Poster in Training · Answer 5 · 2006-07-16T00:12:58+00:00

Sorry, but could you please a little explanation of what your code is supposed to do? I refuse to read all code and try to figure that out.

Thanks in advance,
Eddy

schmidty169 2 Light Poster · Answer 6 · 2006-07-16T21:56:04+00:00

Actually if you look in the main at the end that is it. Fill the tree with the data and spit out a few results.
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;
output
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;
// tree1.traverseinorder();

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Dani AI

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