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BST Implementation in C++
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#include <iostream> | |
using namespace std; | |
struct node { | |
int data; | |
node* left; | |
node* right; | |
node() { | |
left = NULL; | |
right = NULL; | |
} | |
node(int _data) { | |
data = _data; | |
left = NULL; | |
right = NULL; | |
} | |
}; | |
int main(void) { | |
node* tree; | |
} | |
void insertInBst(node** tree, int data) { | |
if ((*tree) == NULL) { | |
*tree = new node(data); | |
return; | |
} | |
else { | |
if (data <= (*tree)->data) { | |
insertInBst(&(*tree)->left, data); | |
} | |
else { | |
insertInBst(&(*tree)->right, data); | |
} | |
} | |
} | |
void deleteFromBst(node** tree, int key) { | |
if ((*tree) == NULL) { | |
return; | |
} | |
else { | |
// Buscar el elemento a eliminar | |
if (key < (*tree)->data) { | |
deleteFromBst(&(*tree)->left, key); | |
} | |
else if(key > (*tree)->data) { | |
deleteFromBst(&(*tree)->right, key); | |
} | |
else { | |
// Eliminacion | |
// 1. Si el nodo es hoja (0 hijos) | |
if ((*tree)->left == NULL && (*tree)->right == NULL) | |
free(*tree); | |
// 2. Si el nodo tiene 1 hijo | |
// 2.1 Si el hijo es hijo izquierdo | |
if ((*tree)->left != NULL && (*tree)->right == NULL) { | |
*tree = (*tree)->left; | |
} | |
// 2.2 | |
else if((*tree)->left == NULL && (*tree)->right != NULL) { | |
*tree = (*tree)->right; | |
} | |
// 3. El nodo tiene 2 hijos | |
else { | |
int inOrderSuccesor = getInOrderSuccesor((*tree)->right); | |
(*tree)->data = inOrderSuccesor; | |
deleteFromBst(&(*tree)->right, inOrderSuccesor); | |
} | |
} | |
} | |
} | |
int getInOrderSuccesor(node* tree) { | |
while(tree->left != NULL) { | |
tree = tree->left; | |
} | |
return tree->data; | |
} |
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