[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.
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 return mySortedListToBST(head, getLength(head)); 21 } 22 23 private TreeNode mySortedListToBST(ListNode head, int length) { 24 // TODO Auto-generated method stub 25 if (head == null || length == 0) 26 return null; 27 if (length == 1) 28 return new TreeNode(head.val); 29 int mid = length / 2; 30 ListNode peak = head; 31 for (int i = 0; i < mid; ++i) { 32 peak = peak.next; 33 } 34 TreeNode root=new TreeNode(peak.val); 35 root.left=mySortedListToBST(head, mid); 36 root.right=mySortedListToBST(peak.next, length-mid-1); 37 return root; 38 } 39 40 private int getLength(ListNode head) { 41 // TODO Auto-generated method stub 42 int length = 0; 43 while (head != null) { 44 head = head.next; 45 length++; 46 } 47 return length; 48 } 49 }
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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
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 ListNode cur; 20 public TreeNode sortedListToBST(ListNode head) { 21 if(head==null) 22 return null; 23 ListNode runner=head; 24 int size=1; 25 cur=head; 26 while(runner.next!=null){ 27 runner=runner.next; 28 size++; 29 } 30 return makeTree(0,size-1); 31 } 32 public TreeNode makeTree(int start,int end){ 33 if(start>end) 34 return null; 35 if(start==end){ 36 TreeNode temp= new TreeNode(cur.val); 37 cur=cur.next; 38 return temp; 39 } 40 int mid=start+(end-start)/2; 41 TreeNode left=makeTree(start,mid-1); 42 TreeNode root=new TreeNode(cur.val); 43 root.left=left; 44 cur=cur.next; 45 TreeNode right=makeTree(mid+1,end); 46 root.right=right; 47 return root; 48 } 49 }
