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# simple question about B tree

 P: n/a Hi I am relearning datastructure... but got kinda stuck in a basic delete node operation. what it does is to delete the first node it finds when the item is equal the input item. the end result is that the node is deleted and the b-tree is still kept as b-tree. CAn someone critique on my implementation? Any suggestion for improvement? Thanks The code I have as follows: #include using namespace std; class Btree; class Node { private: int value; Node* left; Node* right; friend Btree; public: Node(int item): value(item),left(0),right(0){}; ~Node(){cout<<"value is :"<left) &&(!cur->right) } } //Helper functions int Btree::swap(Node* node1, Node* node2){ int ret=0; if(!node1 || !node2){ ret =0;} else { int temp = node2->value; node2->value=node1->value; node1->value=temp; ret =1; } return ret; }; #include using namespace std; class Btree; class Node { private: int value; Node* left; Node* right; friend Btree; public: Node(int item): value(item),left(0),right(0){}; ~Node(){cout<<"value is :"<left) &&(!cur->right) } } int Btree::swap(Node* node1, Node* node2){ int ret=0; if(!node1 || !node2){ ret =0;} else { int temp = node2->value; node2->value=node1->value; node1->value=temp; ret =1; } return ret; }; int Btree::removeNode(Node* cur, int item){ int ret=0; if(!cur) { ret =0;} else { if(cur->value == item) { /*get rid of node*/ // if there is a left child // keep promoting the left child up as long as it has one. if (isLeafNode(cur)){ delete cur;} // if I am leaf node delete myself else { // do I have a left node? yes promote, // get rid of the last one in chain while (cur->left) { swap(cur,cur->left); cur=cur->left; } while (cur->right) { swap(cur,cur->right); cur=cur->right; } //should be leaf node pointed by cur if(isLeafNode(cur)){delete cur;} // no promote right node // get rid of the last one in chain } } else{ //if item < cur->value, tell left to do the deletion, // else tell right if (itemvalue) removeNode(cur->left,item); else removeNode(cur->right,item); } } return ret; } Nov 28 '06 #1
3 Replies

 P: n/a as*********@gmail.com wrote: Hi I am relearning datastructure... but got kinda stuck in a basic delete node operation. what it does is to delete the first node it finds when the item is equal the input item. the end result is that the node is deleted and the b-tree is still kept as b-tree. CAn someone critique on my implementation? Any suggestion for improvement? Thanks The code I have as follows: #include using namespace std; class Btree; class Node { private: int value; Node* left; Node* right; friend Btree; public: Node(int item): value(item),left(0),right(0){}; ~Node(){cout<<"value is :"<left) &&(!cur->right) } } //Helper functions int Btree::swap(Node* node1, Node* node2){ int ret=0; if(!node1 || !node2){ ret =0;} else { int temp = node2->value; node2->value=node1->value; node1->value=temp; ret =1; } return ret; }; #include using namespace std; class Btree; class Node { private: int value; Node* left; Node* right; friend Btree; public: Node(int item): value(item),left(0),right(0){}; ~Node(){cout<<"value is :"<left) &&(!cur->right) } } int Btree::swap(Node* node1, Node* node2){ int ret=0; if(!node1 || !node2){ ret =0;} else { int temp = node2->value; node2->value=node1->value; node1->value=temp; ret =1; } return ret; }; int Btree::removeNode(Node* cur, int item){ int ret=0; if(!cur) { ret =0;} else { if(cur->value == item) { /*get rid of node*/ // if there is a left child // keep promoting the left child up as long as it has one. if (isLeafNode(cur)){ delete cur;} // if I am leaf node delete myself else { // do I have a left node? yes promote, // get rid of the last one in chain while (cur->left) { swap(cur,cur->left); cur=cur->left; } while (cur->right) { swap(cur,cur->right); cur=cur->right; } //should be leaf node pointed by cur if(isLeafNode(cur)){delete cur;} // no promote right node // get rid of the last one in chain } } else{ //if item < cur->value, tell left to do the deletion, // else tell right if (itemvalue) removeNode(cur->left,item); else removeNode(cur->right,item); } } return ret; I havent stepped through the code to really see whether it works or not but the first thing that struck me is where are you maintaining the order of the B-tree ? Let's assume you have a b-tree with order 3 and the intermediate nodes have only 2 keys and you delete one of them ...then how to do you maintian the order. your implementation looks more like a binary tree with the adjustment made only for swapping keys at the same level...but not across levels. } Nov 28 '06 #2

 P: n/a I was thinking that since BTree is already sorted. So instead of doing the pointer manipulation, I could just do the following: --- if the item is in a leaf node - delete the node. and then I am done -- if the item is in a node that has 2 children, I swap the value fo the left child (which is always smaller then the parent node --which has the *value* I want to delete, and keep pushing it down until it hits the leftmost leafnode. then I delete the leftmost leaf node. -- if there is no left node but right node, then I push the *value" I want to delete down until it is leaft node then I delete. The core is that because deleting the leaf node is easy. so in order to fill the hole, I need to push down the value down the path until it hits the leaf node. And I also assume that th b-tree is already sorted. do u think it will work? can u give some the implementation? thanks am******@gmail.com wrote: as*********@gmail.com wrote: Hi I am relearning datastructure... but got kinda stuck in a basic delete node operation. what it does is to delete the first node it finds when the item is equal the input item. the end result is that the node is deleted and the b-tree is still kept as b-tree. CAn someone critique on my implementation? Any suggestion for improvement? Thanks The code I have as follows: #include using namespace std; class Btree; class Node { private: int value; Node* left; Node* right; friend Btree; public: Node(int item): value(item),left(0),right(0){}; ~Node(){cout<<"value is :"<left) &&(!cur->right) } } //Helper functions int Btree::swap(Node* node1, Node* node2){ int ret=0; if(!node1 || !node2){ ret =0;} else { int temp = node2->value; node2->value=node1->value; node1->value=temp; ret =1; } return ret; }; #include using namespace std; class Btree; class Node { private: int value; Node* left; Node* right; friend Btree; public: Node(int item): value(item),left(0),right(0){}; ~Node(){cout<<"value is :"<left) &&(!cur->right) } } int Btree::swap(Node* node1, Node* node2){ int ret=0; if(!node1 || !node2){ ret =0;} else { int temp = node2->value; node2->value=node1->value; node1->value=temp; ret =1; } return ret; }; int Btree::removeNode(Node* cur, int item){ int ret=0; if(!cur) { ret =0;} else { if(cur->value == item) { /*get rid of node*/ // if there is a left child // keep promoting the left child up as long as it has one. if (isLeafNode(cur)){ delete cur;} // if I am leaf node delete myself else { // do I have a left node? yes promote, // get rid of the last one in chain while (cur->left) { swap(cur,cur->left); cur=cur->left; } while (cur->right) { swap(cur,cur->right); cur=cur->right; } //should be leaf node pointed by cur if(isLeafNode(cur)){delete cur;} // no promote right node // get rid of the last one in chain } } else{ //if item < cur->value, tell left to do the deletion, // else tell right if (itemvalue) removeNode(cur->left,item); else removeNode(cur->right,item); } } return ret; I havent stepped through the code to really see whether it works or not but the first thing that struck me is where are you maintaining the order of the B-tree ? Let's assume you have a b-tree with order 3 and the intermediate nodes have only 2 keys and you delete one of them ...then how to do you maintian the order. your implementation looks more like a binary tree with the adjustment made only for swapping keys at the same level...but not across levels. } Nov 28 '06 #3

 P: n/a as*********@gmail.com wrote: I was thinking that since BTree is already sorted. Please don't top-post. Your replies belong following or interspersed with properly trimmed quotes. See the majority of other posts in the newsgroup, or the group FAQ list: Nov 28 '06 #4

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