106.Convert Sorted List to Balanced BST-排序列表转换为二分查找树（中等题）

排序列表转换为二分查找树

1. 题目

   给出一个所有元素以升序排序的单链表，将它转换成一棵高度平衡的二分查找树

2. 样例

   

3. 题解

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param head: The first node of linked list.
     * @return: a tree node
     */
    private ListNode current; 

    public TreeNode sortedListToBST(ListNode head) {
        current = head;
        int size = getListLength(head);

        return sortedListToBSTHelper(size);
    }

    private TreeNode sortedListToBSTHelper(int size) 
    {
        if (size <= 0) 
        {
            return null;
        }

        TreeNode left = sortedListToBSTHelper(size / 2);
        TreeNode root = new TreeNode(current.val);
        current = current.next;
        TreeNode right = sortedListToBSTHelper(size - 1 - size / 2);

        root.left = left;
        root.right = right;

        return root;
    }

    private int getListLength(ListNode head) 
    {
        int size = 0;
        while (head != null) 
        {
            size++;
            head = head.next;
        }

        return size;
    }
}

Last Update 2016.10.9