Alphabetically Ordered Binary Tree Word Count Problem

I'm making a binary tree which analyses a text file and counts the frequency of the words from that text file. The problem I have is that I can't seem to increment the count if the the same word has been found again. Any suggestions about what I'm doing wrong would be greatly appreciated!



1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
#include<iostream>
#include<string>
#include<fstream>
#include<algorithm>
#include<iomanip> 

using namespace std;

struct Display
{
	string words;;
	int count;
};

struct Node
{
	Display value;
	Node* pLeft;
	Node* pRight;
};

Node* add(Node* tree, Display value);
Node* addNode(Node* tree, Node* toAdd);

bool treeContains(Node* tree, string word) ;
void displayAlphabetTree(Node* tree); 

void main()
{	
	Display value;
	string choiceStr;
	char choice;
	bool eofReached = false;

	value.count = 0;

	Node* tree = 0;

	string txtFile;

	cout << "Enter filename: ";
	cin >> txtFile;
	cout << endl;

	ifstream read(txtFile.c_str());
	
	if(!read)
	{
		cout << "Error, can't open file" << endl;
		exit(1);
	}
	else
	{		
		while(!read.eof()) 
		{
			read >> value.words;	
			eofReached = read.eof();

			if (treeContains(tree, value.words) == false)
			{				
				tree = add(tree, value);
			}
		}
	}

	read.close();

	do
	{
		cout << "  a - display alphabetic list" << endl;
		cout << "  f - display frequency ordered list" << endl;
		cout << "  s - display statistics" << endl;
		cout << "  x - exit" << endl;
		cout << endl;
		cout << "Enter choice: ";
		cin >> choiceStr;

		choice = choiceStr.length() == 0 ? choice = 'e' : choice = choiceStr[0];

		switch(choice)
		{
		case 'a':
		case 'A':
			displayAlphabetTree(tree);
			break;
		case 'x':
		case 'X':
			break;
		default:
			cout << "Error, invalid choice";
			break;
		}
	}
	while(choice != 'x');
}

Node* add(Node* tree, Display value)
{
        //tempPtr is a temporary pointer

	Node* tempPtr = new Node;
	tempPtr->value = value;
	tempPtr->pLeft  = 0;
	tempPtr->pRight = 0;
	return addNode(tree, tempPtr);
}

Node* addNode(Node* tree, Node* toAdd)
{
	if (tree == 0)
	{
		return toAdd;
	}
	else
	{
		if(toAdd->value.words < tree->value.words)
		{
			tree->pLeft = addNode(tree->pLeft, toAdd);
			return tree;
		}
		else if(toAdd->value.words == tree->value.words)
		{
			tree->value.count++;
			return tree;
		}
		else
		{
			tree->pRight = addNode(tree->pRight, toAdd);
			return tree;

		}
	}

}

bool treeContains(Node* tree, string word)  
{
	// If a word is in the binary
        // tree then true is returned.   
	// Vice versia, false is returned if not.

	if (tree == NULL) 
	{
        // Tree is empty, so it certainly doesn't contain item.
        return false;
    }
   else if (word == tree->value.words) 
   {
       //The word matches to one in the root node.
       return true;
    }
    else if (word < tree->value.words) 
	{
		// The word is less than the one in the root node
		// and must be sent to the left subtree.
		return treeContains(tree->pLeft, word);
    }
    else 
	{
		// The word is more than the one in the root node
		// and must be sent to the right subtree.
        return treeContains(tree->pRight, word);
     }
}


void displayAlphabetTree(Node* tree)
{	
    if(tree != 0)
	{		
		displayAlphabetTree(tree->pLeft);		
		displayAlphabetTree(tree->pRight);	
		cout<<tree->value.words<<"  "<<"["<<tree->value.count<<"]"<<endl;
	}
}
Although your addNode() function handles the case where the word to be added is already found, you
never call it if the word already exists, because addNode() is called by add(), which is called only if
treeContains() returns false (word not found).
Topic archived. No new replies allowed.