[Leetcode] 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.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; next = null; }
 * }
 */
/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        return mySortedListToBST(head, getLength(head));
    }

    private TreeNode mySortedListToBST(ListNode head, int length) {
        // TODO Auto-generated method stub
        if (head == null || length == 0)
            return null;
        if (length == 1)
            return new TreeNode(head.val);
        int mid = length / 2;
        ListNode peak = head;
        for (int i = 0; i < mid; ++i) {
            peak = peak.next;
        }
        TreeNode root=new TreeNode(peak.val);
        root.left=mySortedListToBST(head, mid);
        root.right=mySortedListToBST(peak.next, length-mid-1);
        return root;
    }

    private int getLength(ListNode head) {
        // TODO Auto-generated method stub
        int length = 0;
        while (head != null) {
            head = head.next;
            length++;
        }
        return length;
}
}

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20150215

这个写法（从上到下地来构造这棵树）有点问题：因为现在我无法在O（1）的时间内获取元素，每回都得从头开始遍历一遍，找到中间的才行。因此，我们应该要能够想到得从下到上的来构造这棵树(Each time you are stucked with the top-down approach, give bottom-up a try. )，这样的话，我们能够做到：The bottom-up approach enables us to access the list in its order at the same time as creating nodes.

http://leetcode.com/2010/11/convert-sorted-list-to-balanced-binary.html

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; next = null; }
 * }
 */
/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    ListNode cur;
    public TreeNode sortedListToBST(ListNode head) {
        if(head==null)
            return null;
        ListNode runner=head;
        int size=1;
        cur=head;
        while(runner.next!=null){
            runner=runner.next;
            size++;
        }
        return makeTree(0,size-1);
    }
    public TreeNode makeTree(int start,int end){
        if(start>end)
            return null;
        if(start==end){
            TreeNode temp= new TreeNode(cur.val);
            cur=cur.next;
            return temp;
        }
        int mid=start+(end-start)/2;
        TreeNode left=makeTree(start,mid-1);
        TreeNode root=new TreeNode(cur.val);
        root.left=left;
        cur=cur.next;
        TreeNode right=makeTree(mid+1,end);
        root.right=right;
        return root;
    }
}