109. Convert Sorted List to Binary Search Tree

题目：

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

链接： http://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/

6/7/2017

2ms, 9%

用快慢指针找到中点，递归做左右子树，注意左子树最后一个node的next应该设为null

注意的问题

1. 左子树最后一个node的next应该设为null，第17行

2. 当head == slow也就是左子树为空的时候，不要递归调用，而应该直接设成null，第18行

public class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        if (head == null) {
            return null;
        }
        ListNode fast, slow, prev = head;
        fast = head;
        slow = head;

        while (fast != null && fast.next != null) {
            prev = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        TreeNode node = new TreeNode(slow.val);
        node.right = sortedListToBST(slow.next);
        prev.next = null;
        node.left = (head == slow)? null: sortedListToBST(head);
        return node;
    }
}

别人的答案

in-order traversal + DFS

http://www.cnblogs.com/yrbbest/p/4437320.html

更多讨论

https://discuss.leetcode.com/category/117/convert-sorted-list-to-binary-search-tree