An irritating problem statement, 28ms C++ solution with idea to approach and handling special cases.

• Before continuing here, if in case you are stuck with any particular test case.
If your code can pass these short test cases your code is complete. Atleast this was what I followed.

1. 1
2. 1->2
3. 1->2->3
4. 1->2->3->4->10->110
5. 1->2->3->4->10->110->111
``````class Solution {
public:

/*since there are no array indices this function is used to calculate the middle node given begin and end nodes.
You might see a lot of IF statements in here, how did I get it ?, it's pure trial and error.
*/

/*The Basic idea to calculate the middle node of a linked list is that, have a fast pointer and a slow pointer, start from the beginning, advance fast pointer by 2 nodes then advance slow pointer by 1 node as you can see in the below function.At the end the slow pointer is the one pointing to middle node.
*/
ListNode* middle(ListNode* begin,ListNode* end) {
ListNode* n1=begin, *n2=begin; // n1- slow pointer, n2-fast pointer.
while(n2!=end) {
n2 = n2->next;
if (n2 == end)
return n1;
if (n2->next)//checking for NULL case.
n2 = n2->next;
else
break;
n1 = n1->next;
}
return n1; // return the middle node
}

/*This function is pretty easy to understand ,during each recursive call we need to send beginning node until mid-1 to construct the left subtree to do so, I advance a pointer until my next node is not the middle node and finally returen the node.*/

ListNode* mid_minus_one(ListNode* begin, ListNode*end) {
if (begin == end)
return NULL;
ListNode* n = begin;
while (n->next != end)
n = n->next;
return n;
}

/*Finally here comes the main recursive function, where I construct the left and right subtree recursively*/
TreeNode* construct(ListNode* begin, ListNode* end) {
TreeNode* root = NULL;
if (!begin || !end)
return root;
if (begin == end)
{
root = new TreeNode(begin->val);
return root;
}
ListNode* mid = middle(begin,end);
root = new TreeNode(mid->val);
root->left = construct(begin, mid_minus_one(begin, mid));//left subtree, I pass begin,mid-1
root->right = construct(mid->next, end);//right subtree, I pass mid+1 and end.
return root;
}

return NULL;
while (end -> next)
end = end->next;