Hi, I am doing an assignment about binary search tree. I have problem with remove the root in binary tree. When i remove the root of the binary tree and I view the binary tree in In-Order Traversal, my program will stop running.

``````void myCarSystem::remove(string m)
{
//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->Model.compare(m) == 0)
{
found = true;
break;
}
else
{
parent = curr;
if(m.compare(curr->Model) > 0)
curr = curr->right;
else
curr = curr->left;
}
}
if(!found)
{
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
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->Model = lcurr->Model;

delete lcurr;
lcurrp->left = NULL;
}
else
{
tree_node* tmp;
tmp = curr->right;
curr->Model = tmp->Model;
curr->right = tmp->right;
delete tmp;
}

}
return;
}

}``````

Please give some guild to me. Thank you.
Cjjack88

## All 6 Replies

When i remove the root of the binary tree and I view the binary tree in In-Order Traversal, my program will stop running.

Then you are not resetting the links the way they should be. Removing the root means replacing the root with either the inorder predecessor or successor. Something like this:

``````Original:
D
B     F
A C   E G

Delete the root:
*
B     F
A C   E G

Make the inorder predecessor the root:
C
B     F
A *   E G

Delete the inorder predecessor:
C
B     F
A     E G``````

That behavior should fall out of your deletion algorithm because it is the same as the two child case for any node. But you also need to reset the root pointer. In the example above, if `root` still points to D after the algorithm finishes, the tree will be broken. So at the end, the address of C should be assigned to `root` to finish things up, but only if it was D that was deleted. That is an easy test because `parent` will be `NULL` if the root has a matching value.

commented: nice explaination +1

Thank you for reply, i need some time to understand it. If i just have 3 data in the tree, eg: 1, 2 and 3, and i remove 1 in thee, should 3 become the new root in this tree?

If i just have 3 data in the tree, eg: 1, 2 and 3, and i remove 1 in thee, should 3 become the new root in this tree?

If 1 is the root, then either 2 or 3 could be the new root. You can pick whether the inorder predecessor or successor becomes the new root. But it is not as easy as just making the root's immediate left or right child the new root because you still have to maintain the binary search invariant that an inorder traversal produces a sequence that is sorted in ascending order.

ok, I try to code it 1st....
Thank you..

thankssssssssssss alottttttt

pardon me...what do u mean by, chkr, lcurrp, tmp, curr and lcurr in your program. I am a newbie in programming. But I am willing to know about it.

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