leetcode109~Convert Sorted List to Binary Search Tree

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

/*
* Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
* 给出一个单链表，里面的元素是升序排列。
* 普通解法是把单链表转化成数组，然后采用上题解法。这里不予给出了。
*/

public class ConvertSortedListtoBST109 {    //设置一个快慢指针，快指针每次两步，慢指针每次一步，这样快指针到头的时候，慢指针正好到中间     public TreeNode sortedListToBST2(ListNode head) {            if(head==null) return null;            return helper(head,null);        }    private TreeNode helper(ListNode head, ListNode target) {        if(head == target) return null;        ListNode fast = head;        ListNode slow = head;        while(fast!=target && fast.next!=target) {            fast = fast.next.next;            slow = slow.next;        }        //此时slow到达中间位置        TreeNode root = new TreeNode(slow.val);        root.left = helper(head,slow);        root.right = helper(slow.next,target);        return root;    }    //设置一个全局变量    ListNode curr;    //自底向上建立BST    public TreeNode sortedListToBST(ListNode head) {        curr = head;        int len = 0;        //计算链表的长度        while(head!=null) {            head = head.next;            len++;        }        //建树        return buildTree(0,len-1);    }    private TreeNode buildTree(int start, int end) {        //没有子节点，直接返回        if(start>end) return null;        int mid = start+(end-start)/2; //防止溢出        TreeNode left = buildTree(start, mid-1);        TreeNode root = new TreeNode(curr.val);        //链表顺序遍历        curr = curr.next;        TreeNode right = buildTree(mid+1, end);        //将节点链接起来        root.left = left;        root.right = right;        return root;    }}

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