I gues jalaj's solun works perfect.
Thanks and regards,
Gajendra Dadheech
On Sun, Feb 6, 2011 at 4:06 PM, jalaj <jalaj.jaiswa...@gmail.com> wrote:
> use a stack for this
> {
> //let preorder traversal of a tree be in a array t
> for(i = t.length; i>-1; i--){
>
> if(t(i) == L){
> stack.push(t[i]);
> }else{
> leftChild = s.pop(); // will return null if stack is empty
> rightChild = s.pop(); // will return null if stack is empty
> node = new Node(leftChild, rightChild);
> stack.push(node);
> }
>
> }
>
> root = stack.pop(); // get the root from the stack;
> }
>
> On Feb 6, 3:03 pm, algoseeker <newton.anu...@gmail.com> wrote:
> > I also encountered this question yesterday. This is my solution which i
> > tested for a few sample cases.
> >
> > https://github.com/algoseeker/Interview/blob/master/Node.java
> >
> > I maintained a pointer to the Node where there should be creation of a
> new
> > node. If the node created is left child, shift the pointer to new node
> > created. If this new Node's right child gets created, shift the pointer
> to
> > the parent of this Node.
>
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