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

一刷
题解：利用快慢指针找到median, 快指针每次移动两步， 慢指针每次移动一步。最后慢指针所在位置为median. 并且，这题很自然地用recursion做更方便。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        if(head == null) return null;
        return toBST(head, null);
    }

    
    TreeNode toBST(ListNode head, ListNode tail){
        ListNode slow = head;
        ListNode fast = head;
        if(head == tail) return null;
        
        //the place of slow is the median
        while(fast!=tail && fast.next!=tail){
            fast = fast.next.next;
            slow = slow.next;
        }
        
        TreeNode thread = new TreeNode(slow.val);
        thread.left = toBST(head, slow);
        thread.right = toBST(slow.next, tail);
        return thread;
        
    }
}

二刷：
第二遍做还是忘了可以用快慢指针得到medium的位置。 recursion求解

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        if(head == null) return null;
        return sortedListToBST(head, null); 
    }
    
    private TreeNode sortedListToBST(ListNode head, ListNode tail){
        if(head == null || head == tail) return null;
        ListNode 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 = slow==null? null:sortedListToBST(slow.next, tail);
        return root;
    } 
}

三刷
注意快慢指针的使用

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        if(head == null) return null;
        return toBST(head, null);
    }
    
    public TreeNode toBST(ListNode from, ListNode to){
        //not include null
        if(from == to || from == null) return null;
        ListNode med = findMedium(from, to);
        
        TreeNode root = new TreeNode(med.val);
        root.left = toBST(from, med);
        if(med == null) root.right = null;
        else root.right = toBST(med.next, to);
        return root;
    }
    
    private ListNode findMedium(ListNode head, ListNode tail){
        ListNode slow = head;
        ListNode fast = head;
        while(fast!=tail && fast.next!=tail){
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }
}