06. Convert Sorted List to Binary Search Tree.cpp

/*
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.

Example 1:
Input: head = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: One possible answer is [0,-3,9,-10,null,5], which represents the shown height balanced BST.

Example 2:
Input: head = []
Output: []

Example 3:
Input: head = [0]
Output: [0]

Example 4:
Input: head = [1,3]
Output: [3,1]

Constraints:
The number of nodes in head is in the range [0, 2 * 104].
-10^5 <= Node.val <= 10^5
*/

/**
 * 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
{
public:
  TreeNode *sortedListToBST(ListNode *head, ListNode *tail = NULL)
  {
    if (head == tail)
      return NULL;
    if (head->next == tail)
    {
      TreeNode *root = new TreeNode(head->val);
      return root;
    }

    auto fast = head, slow = head;
    while (fast != tail && fast->next != tail)
    {
      fast = fast->next->next;
      slow = slow->next;
    }

    TreeNode *root = new TreeNode(slow->val);
    root->left = sortedListToBST(head, slow);
    root->right = sortedListToBST(slow->next, tail);

    return root;
  }
};