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/*** 题目：* Given 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.* 解题思路：* 利用快慢指针找到中间节点，作为根节点，然后左子树即为左边链表部分，右子树即为右边链表部分，递归进行即可。*/
public class SortedListToBST_109_1016 {public TreeNode SortedListToBST(ListNode head) {if (head == null) {return null;}return BST(head, null);}public TreeNode BST(ListNode head, ListNode tail) {if (head == tail) {return null;}//利用快慢指针找到中间节点，作为根节点ListNode slow = head;ListNode fast = head;while (fast != tail && fast.next != tail) {fast = fast.next.next;slow = slow.next;}TreeNode root = new TreeNode(slow.val);root.left = BST(head, slow);root.right = BST(slow.next, tail);return root;}
}
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