Leetcode-Convert Sorted List to BST.

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; next = null; }
 7  * }
 8  */
 9 /**
10  * Definition for binary tree
11  * public class TreeNode {
12  *     int val;
13  *     TreeNode left;
14  *     TreeNode right;
15  *     TreeNode(int x) { val = x; }
16  * }
17  */
18 public class Solution {
19     public TreeNode sortedListToBST(ListNode head) {
20         if (head==null)
21             return null;
22         
23         ListNode end = head;
24         while (end.next!=null){
25             end = end.next;
26         }
27         TreeNode root = sortedListToBSTRecur(head,end);
28         
29         return root;
30     }
31     
32     //Convert the list from head to end into BST, return the root node
33     public TreeNode sortedListToBSTRecur(ListNode head, ListNode end){
34         //If only one node.
35         if (head==end){
36             TreeNode root = new TreeNode(head.val);
37             return root;
38         }
39         
40         //If only two node.
41         if (head.next==end){
42             TreeNode root = new TreeNode(head.val);
43             TreeNode child = new TreeNode(end.val);
44             root.right = child;
45             return root;
46         }
47         
48         
49         //Otherwise, count the number of node in the list, find out the median one, 
50         //and convert the left list and right list to BST.
51         int nodeNum = 1;
52         ListNode curNode = head;
53         while (curNode!=end){
54             curNode = curNode.next;
55             nodeNum++;
56         }
57         int mid = nodeNum/2+nodeNum%2;
58         ListNode median = null;
59         ListNode pre = null;
60         curNode = head;
61         //NOTE: we only need to move (mid-1) steps to move curNode to the median one.
62         for (int i=0;i<mid-1;i++){
63             pre = curNode;
64             curNode = curNode.next;
65         }
66         median = curNode;
67         TreeNode root = new TreeNode(median.val);
68         TreeNode leftChild = sortedListToBSTRecur(head,pre);
69         TreeNode rightChild = sortedListToBSTRecur(median.next,end);
70         root.left = leftChild;
71         root.right = rightChild;
72         return root;
73     }
74 }