109. Convert Sorted List to Binary Search Tree

Given the head of a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
// n / 2 + 2 * n / 4 + 4 * n / 8 .... a * n / n
// n / 2 + n / 2 + n / 2 + .... a * n / n
// n / 2 * log n = n log n is time 
public:
    TreeNode *sortedListToBST(ListNode *head) {
        if(head == nullptr) return nullptr;
        if(head->next == nullptr) return new TreeNode(head->val);

        ListNode *slow = head, *fast = head;
        while(fast->next->next and fast->next->next->next) {
            fast = fast->next->next;
            slow = slow->next;
        }
        ListNode* mid = slow->next;
        ListNode* right = mid->next;
        ListNode* left = head;
        slow->next = nullptr;
        mid->next = nullptr;

        TreeNode* node = new TreeNode(mid->val);
        node->left = sortedListToBST(left);
        node->right = sortedListToBST(right);
        return node;
    }
};