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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