A binary search tree is defined as follows.
Each node has a key, a value, a (possibly null) left child, and a
(possibly null) right child.
An ordering is defined on the keys to be stored in the tree.
For a given node nd, the keys of all the nodes in its left subtree are
“before” nd’s key, and the keys of all the nodes in its right subtree are
“after” nd’s key.
The depth of the root node is defined to be 0; the depth of any other
node is 1 more than the depth of its parent.
How would the next() function of a binary search tree’s iterator be defined,
assuming no changes are made to the tree while the iterator is active?
Note: next() should return the value at each node in order of the
corresponding keys; also, assume for each answer that the iterator has been
appropriately initialized before the first call to next().
1. using a queue:
nd = q.remove()
q.add(nd.leftChild)
a.qdd(nd.rightChild)
return nd.value
2. using a stack:
nd = stk.remove()
stk.add(nd.leftChild)
stk.add(nd.rightChild)
return nd.value
3. using a stack:
nd = stk.remove()
nxt = nd.getRight()
while (null != nxt) { stk.push(nxt); nxt = nxt.getLeft(); }
return nd.value
1 is correct
2 is correct
3 is correct
None of these
plz say ur answers for this ??
On Fri, Aug 19, 2011 at 9:18 PM, jestincobol nair <jestinco...@gmail.com>wrote:
> @ sagar:no dude !! dese are of not of citrix :)
>
>
> On Fri, Aug 19, 2011 at 9:14 PM, sagar pareek <sagarpar...@gmail.com>wrote:
>
>> @nair
>> Are these questions of CITRIX?
>>
>> coz all these questions i already faced somewhere
>>
>>
>> On Fri, Aug 19, 2011 at 9:12 PM, priya ramesh <
>> love.for.programm...@gmail.com> wrote:
>>
>>> 5th is option 1 sorry
>>>
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>>
>>
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>> **Regards
>> SAGAR PAREEK
>> COMPUTER SCIENCE AND ENGINEERING
>> NIT ALLAHABAD
>>
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