Binary Search Tree - unique values in C++
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-4
The project is to add words to the to each nodes in the BST.
I need to count the number of unique or distinct values in my BST.
Here is my code for Adding the words. I need help with writing
int distinctWords() const;.
void WordTree:: addPrivate(WordNode *n, ItemType v)
{
if (root == NULL)
root = new WordNode(v);
else if (v == n->m_data)
{
n->m_count++;
}
else if (v < n->m_data)
{
if (n->m_left != NULL)
{
addPrivate(n->m_left, v);
}
else
{
n->m_left = new WordNode(v);
}
}
else if (v > n->m_data)
{
if (n->m_right != NULL)
{
addPrivate(n->m_right, v);
}
else
{
n->m_right = new WordNode(v);
}
}
}
c++ binary-search-tree
asked Nov 23 '18 at 21:02
Artin Artin
11
add a comment |
-4
The project is to add words to the to each nodes in the BST.
I need to count the number of unique or distinct values in my BST.
Here is my code for Adding the words. I need help with writing
int distinctWords() const;.
void WordTree:: addPrivate(WordNode *n, ItemType v)
{
if (root == NULL)
root = new WordNode(v);
else if (v == n->m_data)
{
n->m_count++;
}
else if (v < n->m_data)
{
if (n->m_left != NULL)
{
addPrivate(n->m_left, v);
}
else
{
n->m_left = new WordNode(v);
}
}
else if (v > n->m_data)
{
if (n->m_right != NULL)
{
addPrivate(n->m_right, v);
}
else
{
n->m_right = new WordNode(v);
}
}
}
c++ binary-search-tree
asked Nov 23 '18 at 21:02
Artin Artin
11
You have 2 ways: travel the tree to compute the value, or store it as member and updating the value with insertion/removal of word.
– Jarod42
Nov 23 '18 at 21:06
add a comment |
-4
-4
-4
The project is to add words to the to each nodes in the BST.
I need to count the number of unique or distinct values in my BST.
Here is my code for Adding the words. I need help with writing
int distinctWords() const;.
void WordTree:: addPrivate(WordNode *n, ItemType v)
{
if (root == NULL)
root = new WordNode(v);
else if (v == n->m_data)
{
n->m_count++;
}
else if (v < n->m_data)
{
if (n->m_left != NULL)
{
addPrivate(n->m_left, v);
}
else
{
n->m_left = new WordNode(v);
}
}
else if (v > n->m_data)
{
if (n->m_right != NULL)
{
addPrivate(n->m_right, v);
}
else
{
n->m_right = new WordNode(v);
}
}
}
c++ binary-search-tree
asked Nov 23 '18 at 21:02
Artin Artin
11
The project is to add words to the to each nodes in the BST.
I need to count the number of unique or distinct values in my BST.
Here is my code for Adding the words. I need help with writing
int distinctWords() const;.
void WordTree:: addPrivate(WordNode *n, ItemType v)
{
if (root == NULL)
root = new WordNode(v);
else if (v == n->m_data)
{
n->m_count++;
}
else if (v < n->m_data)
{
if (n->m_left != NULL)
{
addPrivate(n->m_left, v);
}
else
{
n->m_left = new WordNode(v);
}
}
else if (v > n->m_data)
{
if (n->m_right != NULL)
{
addPrivate(n->m_right, v);
}
else
{
n->m_right = new WordNode(v);
}
}
}
c++ binary-search-tree
c++ binary-search-tree
asked Nov 23 '18 at 21:02
Artin Artin
11
asked Nov 23 '18 at 21:02
Artin Artin
11
asked Nov 23 '18 at 21:02
Artin Artin
11
asked Nov 23 '18 at 21:02
Artin Artin
11
asked Nov 23 '18 at 21:02
Artin Artin
11
11
You have 2 ways: travel the tree to compute the value, or store it as member and updating the value with insertion/removal of word.
– Jarod42
Nov 23 '18 at 21:06
add a comment |
You have 2 ways: travel the tree to compute the value, or store it as member and updating the value with insertion/removal of word.
– Jarod42
Nov 23 '18 at 21:06
You have 2 ways: travel the tree to compute the value, or store it as member and updating the value with insertion/removal of word.
– Jarod42
Nov 23 '18 at 21:06
You have 2 ways: travel the tree to compute the value, or store it as member and updating the value with insertion/removal of word.
– Jarod42
Nov 23 '18 at 21:06
add a comment |
1 Answer
1
active
oldest
votes
0
With that tree, that's the same as the number of nodes.
Determining the number of nodes recursively:
- If the tree is empty, it has no nodes.
- If it's not empty, it has one more node than the sum of the number of nodes in the subtrees.
In C++,
int distinctWords(const WordNode* node)
{
return node == nullptr
? 0
: 1 + distinctWords(node->m_left) + distinctWords(node->m_right);
}
answered Nov 23 '18 at 21:13
molbdnilomolbdnilo
41.8k32152
You currently also count duplicated word, checkingn->m_count...
– Jarod42
Nov 23 '18 at 21:46
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
With that tree, that's the same as the number of nodes.
Determining the number of nodes recursively:
- If the tree is empty, it has no nodes.
- If it's not empty, it has one more node than the sum of the number of nodes in the subtrees.
In C++,
int distinctWords(const WordNode* node)
{
return node == nullptr
? 0
: 1 + distinctWords(node->m_left) + distinctWords(node->m_right);
}
answered Nov 23 '18 at 21:13
molbdnilomolbdnilo
41.8k32152
You currently also count duplicated word, checkingn->m_count...
– Jarod42
Nov 23 '18 at 21:46
add a comment |
0
With that tree, that's the same as the number of nodes.
Determining the number of nodes recursively:
- If the tree is empty, it has no nodes.
- If it's not empty, it has one more node than the sum of the number of nodes in the subtrees.
In C++,
int distinctWords(const WordNode* node)
{
return node == nullptr
? 0
: 1 + distinctWords(node->m_left) + distinctWords(node->m_right);
}
answered Nov 23 '18 at 21:13
molbdnilomolbdnilo
41.8k32152
You currently also count duplicated word, checkingn->m_count...
– Jarod42
Nov 23 '18 at 21:46
add a comment |
0
0
0
With that tree, that's the same as the number of nodes.
Determining the number of nodes recursively:
- If the tree is empty, it has no nodes.
- If it's not empty, it has one more node than the sum of the number of nodes in the subtrees.
In C++,
int distinctWords(const WordNode* node)
{
return node == nullptr
? 0
: 1 + distinctWords(node->m_left) + distinctWords(node->m_right);
}
answered Nov 23 '18 at 21:13
molbdnilomolbdnilo
41.8k32152
With that tree, that's the same as the number of nodes.
Determining the number of nodes recursively:
- If the tree is empty, it has no nodes.
- If it's not empty, it has one more node than the sum of the number of nodes in the subtrees.
In C++,
int distinctWords(const WordNode* node)
{
return node == nullptr
? 0
: 1 + distinctWords(node->m_left) + distinctWords(node->m_right);
}
answered Nov 23 '18 at 21:13
molbdnilomolbdnilo
41.8k32152
answered Nov 23 '18 at 21:13
molbdnilomolbdnilo
41.8k32152
answered Nov 23 '18 at 21:13
molbdnilomolbdnilo
41.8k32152
answered Nov 23 '18 at 21:13
molbdnilomolbdnilo
41.8k32152
41.8k32152
You currently also count duplicated word, checkingn->m_count...
– Jarod42
Nov 23 '18 at 21:46
add a comment |
You currently also count duplicated word, checkingn->m_count...
– Jarod42
Nov 23 '18 at 21:46
You currently also count duplicated word, checking
n->m_count...– Jarod42
Nov 23 '18 at 21:46
You currently also count duplicated word, checking
n->m_count...– Jarod42
Nov 23 '18 at 21:46
add a comment |
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You have 2 ways: travel the tree to compute the value, or store it as member and updating the value with insertion/removal of word.
– Jarod42
Nov 23 '18 at 21:06