109. Convert Sorted List to Balanced BST

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

Example

               2
1->2->3  =>   / \
             1   3

分析：
非常简单，用递归即可。需要注意返回mid node的时候，要把整个list分成两半。

public class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        return helper(head, null);
    }

    public TreeNode helper(ListNode head, ListNode tail) {
        ListNode mid = middle(head, tail);
        if (mid == null) return null;
        TreeNode root = new TreeNode(mid.val);
        root.right = helper(mid.next, tail);
        root.left = helper(head, mid);
        return root;
    }

    private ListNode middle(ListNode head, ListNode tail) {
        if (head == tail) return null;
        ListNode slow = head, quick = head;
        while (quick != tail && quick.next != tail) {
            slow = slow.next;
            quick = quick.next.next;
        }
        return slow;
    }
}

 

/**
 * 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;
        ListNode mid = middle(head);
        TreeNode root = new TreeNode(mid.val);
        root.right = sortedListToBST(mid.next);
        if (mid != head) {
            root.left = sortedListToBST(head);
        }
        return root;
    }
    
    private ListNode middle(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode pre = null, slow = head, quick = head;
        
        while(quick.next != null && quick.next.next != null) {
            pre = slow;
            slow = slow.next;
            quick = quick.next.next;
        }
        
        if (pre != null) {
            pre.next = null;  // cut the list into halves.
        }
        return slow;
    }
}