5.3.5 Convert Sorted List to Binary Sear Tree

Notes:
	Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
	Solution: 1. Recursion. Pre-order. A very good Idea. O(n)
	2. Recursion, Binary Search.
	*/
	/**
	* Definition for singly-linked list.
	* struct ListNode {
	* int val;
	* ListNode *next;
	* ListNode(int x) : val(x), next(NULL) {}
	* };
	*/
	/**
	* Definition for binary tree
	* struct TreeNode {
	* int val;
	* TreeNode *left;
	* TreeNode *right;
	* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
	* };

class Solution {
public:
    TreeNode *sortedListToBST_1(ListNode *head) {
        return sortedListToBSTRe(head, getLength(head));
    }
    
    TreeNode *sortedListToBSTRe(ListNode *&head, int length)
    {
        if (length == 0) return NULL;
        int mid = length / 2;
        TreeNode *left = sortedListToBSTRe(head, mid);
        TreeNode *root = new TreeNode(head->val);
        TreeNode *right = sortedListToBSTRe(head->next, length - mid - 1);
        root->left = left;
        root->right = right;
        head = head->next;
        return root;
    }
    
    int getLength(ListNode *head)
    {
        int length = 0;
        while (head)
        {
            length++;
            head = head->next;
        }
        return length;
    }

    TreeNode *sortedListToBST_2(ListNode *head) {
        if(head==nullptr) return nullptr;
        if(head->next==nullptr) return new TreeNode(head->val);
        ListNode * slow = head;
        ListNode * fast = head;
        ListNode * pre = nullptr;
        while(fast->next&&fast->next->next){
            pre = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        fast = slow->next;
        TreeNode * node = new TreeNode(slow->val);
        if(pre){
            pre->next = nullptr;
            node->left = sortedListToBST(head);
        }
        node->right = sortedListToBST(fast);
        return node;
    }
};