[英]Remove fcn for Binary Search Tree
二進制搜索樹的此刪除功能看起來正確嗎? 當我嘗試刪除節點時,它會運行並帶回它從中調用的切換菜單,但是當您重新打印樹時實際上並沒有刪除它。 我不確定是否有退貨的地方,可能嗎?
我的問題是:此remove語句實際上是刪除通過的項目,還是只是以殘酷的方式玩弄它而使我頭疼?
void BinarySearchTree::remove(int d)
{
//Locate the element
bool found = false;
if(isEmpty())
{
cout<<" This Tree is empty! "<<endl;
return;
}
tree_node* curr;
tree_node* parent;
curr = root;
while(curr != NULL)
{
if(curr->data == d)
{
found = true;
break;
}
else
{
parent = curr;
if(d>curr->data) curr = curr->right;
else curr = curr->left;
}
}
if(!found)
{
cout<<" Data not found! "<<endl;
return;
}
// 3 cases :
// 1. We're removing a leaf node
// 2. We're removing a node with a single child
// 3. we're removing a node with 2 children
// Node with single child
// Node with single child
if((curr->left == NULL && curr->right != NULL) || (curr->left != NULL && curr->right == NULL))
{
if(curr->left == NULL && curr->right != NULL)
{
if(parent->left == curr)
{
parent->left = curr->right;
delete curr;
}
else
{
parent->right = curr->right;
delete curr;
}
}
else // left child present, no right child
{
if(parent->left == curr)
{
parent->left = curr->left;
delete curr;
}
else
{
parent->right = curr->left;
delete curr;
}
}
return;
}
//We're looking at a leaf node
if( curr->left == NULL && curr->right == NULL)
{
if(parent->left == curr)
{
parent->left = NULL;
}
else
{
parent->right = NULL;
}
delete curr;
return;
}
//Node with 2 children
// replace node with smallest value in right subtree
if (curr->left != NULL && curr->right != NULL)
{
tree_node* chkr;
chkr = curr->right;
if((chkr->left == NULL) && (chkr->right == NULL))
{
curr = chkr;
delete chkr;
curr->right = NULL;
}
else // right child has children
{
//if the node's right child has a left child
// Move all the way down left to locate smallest element
if((curr->right)->left != NULL)
{
tree_node* lcurr;
tree_node* lcurrp;
lcurrp = curr->right;
lcurr = (curr->right)->left;
while(lcurr->left != NULL)
{
lcurrp = lcurr;
lcurr = lcurr->left;
}
curr->data = lcurr->data;
delete lcurr;
lcurrp->left = NULL;
}
else
{
tree_node* tmp;
tmp = curr->right;
curr->data = tmp->data;
curr->right = tmp->right;
delete tmp;
}
}
return;
}
}
我讀了您的代碼,並在紙上寫了不同的情況,我認為在這種情況下,該節點有2個孩子,代碼應該是這樣的
//Node with 2 children
if (curr->left != NULL && curr->right != NULL)
{
tree_node* chkr;
if(parent==NULL || parent->left==curr){ //if parent==NULL it means that the node that we want to delete is root
chkr=curr->right;
while(chkr->left!=NULL)
chkr=chkr->left;
if(parent!=NULL)
parent->left=curr->right;
else
root=curr->right;
chkr->left=curr->left;
curr->left=curr->right=NULL;
delete curr;
}else if(parent->right==curr){
chkr=curr->left;
while(chkr->right!=NULL)
chkr=chkr->right;
parent->right=curr->left;
chkr->right=curr->right;
curr->left=curr->right=NULL;
delete curr;
}
return;
}
我沒有對其進行編譯或測試,但我認為它會起作用,請讓我知道
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