Convert Sorted Array to Binary Search Tree

# Definition for a  binary tree node
class TreeNode:
	def __init__(self, x):
		self.val = x
		self.left = None
		self.right = None
	def disp(self):
		print self.val
		if(self.left!=None):
			self.left.disp()
		if(self.right!=None):
			self.right.disp()
# Definition for singly-linked list.
class ListNode:
	def __init__(self, x):
		self.val = x
		self.next = None

class Solution:
    # @param head, a list node
    # @return a tree node
    def sortedListToBST(self, head):
    	l=[]
    	while(head!=None):
    		l.append(head.val)
    		head=head.next
    	#print l
    	return self.sortArray2BST(l)
    def sortArray2BST(self,arr):
    	if(len(arr)==0):
    		return None
    	if(len(arr)==1):
    		return TreeNode(arr[0])
    	else:
    		mid=len(arr)/2
    		treenode=TreeNode(arr[mid])
    		treenode.left=self.sortArray2BST(arr[0:mid])
    		treenode.right=self.sortArray2BST(arr[mid+1:len(arr)])
    		return treenode

将linked-list转化为array，有别人提出的不转化的办法是通过两个指针，第一个指针每次走一位，第二个指针每次走两位，第二个指针走到最后的时候，第一个指针走到中间位。
可用多个指针（速度不同）进行对链表的定位