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Microsoft Interview Question


Country: India
Interview Type: In-Person
More Questions from This Interview



Comment hidden because of low score. Click to expand.
1
of 1 vote

delete the nodes in postorder traversal.

- Anonymous June 30, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

You can watch and understand the code to delete all the nodes of a binary search tree from this amazing video tutorial
youtu.be/qZbmY-2Y26Y

- Anonymous June 30, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

Good effort. Answer is: Post Order traversal and delete the node where you would usually print it

- Sandeep July 07, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

hey your ques are coming couple of times .... please remove the duplicate ones.... thanks a lot

- student June 30, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

I am very sorry.. I have clicked submit button twice. But now I am unable delete the duplicate questions.

- gdb June 30, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

also where all this ques in written test or interview....

- student June 30, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

delete the nodes in post order traversal..

- Anonymous June 30, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void main(String args[])
{
  Read Tree root T
  DeleteNodeInPostOrder(T)
  T = null;
}
public static void DeleteNodeInPostOrder(Node T)
{
  if(T == null)
    return;
  DeleteNodeInPostOrder(T.left);
  T.Left = null;
  DeleteNodeInPostOrder(T->right);
  T.right = null;
}

- Randomness June 30, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

Where are your nodes deleted?
You are simply setting them to NULL. Effectively leaking memory.

- Pavitra.Ghosh July 11, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

void deleteBst(struct node** node){
     if (*node == NULL)
        return;
    else if ( !((*node)->left) && !( (*node)->right) )) {
        free(*node);
        *node=NULL;
    }
    else {
       deleteBst( &((*node)->left)));
       deleteBst( &((*node)->right)));
    }

}

- coolsolution June 30, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

Sorry, last solution is little wrong this is the correct working code

void deleteBst(struct node** node){
    
     if (*node == NULL)
        return;
    
     printf("traversing %d node now\n",(*node)->data);
     
     deleteBst( &((*node)->left));
     deleteBst( &((*node)->right));
     
     if ( !((*node)->left) && !( (*node)->right) ) {
        printf("deleting node %d now\n",(*node)->data);
        free(*node);
        *node=NULL;
     }
    
    
}

- coolsolution June 30, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

You can watch learn how to delete all the nodes of a binary search tree from this wonderful video tutorial
youtu.be/qZbmY-2Y26Y

- Anonymous June 30, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

In your code, do we actually need the condition to free the node..? Won`t it work without the condition..? Please explain me if I am wrong..!

if ( !((*node)->left) && !( (*node)->right) ) {
   // free the node
}

- f2003062 July 16, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

Note: After deleting the BST make root=NULL

void deleteBST(root)
{
	if(root)
	{
		deleteBST(root->left);
		deleteBST(root->right);

		free(root);
                root=NULL;
	}
}

- Aashish June 30, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

{{
delete_bst(bst_node *root) {
if (root) {
delete_bst(root->left);
delete_bst(root->right);
delete(root);
}
}}

- xx June 30, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

test(){

   }

- test June 30, 2012 | Flag Reply
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0
of 0 vote

Non recursive approach could be DFT (Depth First Search) and delete nodes from top of stack...

- TS July 02, 2012 | Flag Reply
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0
of 0 vote

Can we not do it using inorder, preorder traversals too?

- Anonymous July 05, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

Post order traversal only works because while deleting the nodes, first delete the childs (i.e. left and right nodes) and then delete the root itself..

- f2003062 July 16, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

Perform BFS-

void modified_bfs(node *root)
{
	if( root == 0 )
		return;
	q.add(root);
	while( !q.empty() )
	{
		node* x = q.delete();
		if( x->left )
			q.add(x->left);
		if( x->right )
			q.add(x->right)
		delete x; /*free x */
	}
}

- confused_banda December 25, 2013 | Flag Reply


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