【LeetCode】C# 109、Convert Sorted List to Binary Search Tree

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

题意：给定一个升序单向链表，将其转化为以一个高度平衡的查询二叉树。

思路：跟上题一样，由递归思路每次找中数作为根，其左右边的链表分别作为左右子树。多出来的问题在于怎样找到中位数。可以通过新建slow、fast两个节点遍历一次链表段来找出。

/**
* Definition for singly-linked list.
* public class ListNode {
*     public int val;
*     public ListNode next;
*     public ListNode(int x) { val = x; }
* }
*/
/**
* Definition for a binary tree node.
* public class TreeNode {
*     public int val;
*     public TreeNode left;
*     public TreeNode right;
*     public 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 tree,ListNode list){
ListNode slow = tree;
ListNode fast = tree;
if(tree==list) return null;
while(fast!=list&&fast.next!=list){
fast = fast.next.next;
slow = slow.next;
}
TreeNode thead = new TreeNode(slow.val);
thead.left = toBST(tree,slow);
thead.right = toBST(slow.next,list);
return thead;
}
}