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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.
解法一：双节点
public TreeNode sortedListToBST(ListNode head) { if(head == null) return null; ListNode fast = head; ListNode slow = head; ListNode prev =null; while(fast != null && fast.next != null) { fast = fast.next.next; prev =slow; slow=slow.next; } TreeNode root = new TreeNode(slow.val); if(prev != null) prev.next = null; //断掉前面序列和中点之间的联系，不然recursion没法找到前面序列的中点 else head = null; root.left = sortedListToBST(head); root.right = sortedListToBST(slow.next); return root; }
reference：https://leetcode.com/discuss/58428/recursive-construction-using-slow-fast-traversal-linked-list
1 public class Solution { 2 private ListNode current; 3 4 private int getListLength(ListNode head) { 5 int size = 0; 6 7 while (head != null) { 8 size++; 9 head = head.next; 10 } 11 12 return size; 13 } 14 15 public TreeNode sortedListToBST(ListNode head) { 16 int size; 17 18 current = head; 19 size = getListLength(head); 20 21 return sortedListToBSTHelper(size); 22 } 23 24 public TreeNode sortedListToBSTHelper(int size) { 25 if (size <= 0) { 26 return null; 27 } 28 29 TreeNode left = sortedListToBSTHelper(size / 2); 30 TreeNode root = new TreeNode(current.val); 31 current = current.next; 32 TreeNode right = sortedListToBSTHelper(size - 1 - size / 2); 33 34 root.left = left; 35 root.right = right; 36 37 return root; 38 } 39 }
Convert Sorted List to Binary Search Tree
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原文地址：http://www.cnblogs.com/hygeia/p/5058596.html
