Sorry about the tree, didn't keep in mind about BST while drawing that.
But I think you all got the point that I was trying to point out.
On 13 August 2011 23:18, rajul jain rajuljain...@gmail.com wrote:
see WgpShashank second point carefully
it say successor is parent of left node
so
@ashmantak: the figure of dipankar is incorrect but his point is correct..
for a tree like
4
/\
2 10
/ \
13
successor of 3 shall be 4 not 2..
On Fri, Aug 12, 2011 at 4:48 PM, ashmantak
It is the solution for inorder successor without having parent pointer..
node *inorder_succ(node *root, int x)
{
node *prev=NULL;
node *p=findspcl(root,prev, x);
if(p-r)
{
p=p-r;
while(p-l)
p=p-l;
return p;
...
if(something)
return true;
else
return false;
.
// few statements here
.
Will few statements ever execute??
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see WgpShashank second point carefully
it say successor is parent of left node
so solution of you BST is parent of 2 which is 4
On Sat, Aug 13, 2011 at 7:50 PM, Anika Jain anika.jai...@gmail.com wrote:
@ashmantak: the figure of dipankar is incorrect but his point is correct..
for a tree like
void find(struct node * tree,struct node *subtree)
{
if((tree==NULL )||(subtree==NULL))
return ;
else
{
printf(the data being compared are tree:%d subtree:%d
\n,tree-data,subtree-data);
if((tree-data)==(subtree-data))
{
flag=1;
find(tree-left,subtree-left);
find(tree-right,subtree-right);
}
@priya here is algorithm
Algorithm: (With Using Parent Pointer )
1) If right subtree of node is not NULL, then successor lies in right
subtree. Do following.
Go to right subtree and return the node with minimum key value in right
subtree.
2) If right sbtree of node is NULL, then successor is
@Dipankar Patro -
The figure u have made isn't a BST.Read the problem.
His soln. is correct.
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