题意：把有序单链表转换为二叉查找树
思路：
用链表中心作为当作二叉树的根，
链表的前半段作为左子树、后半段作为右子树。
可用递归实现
复杂度：时间O(n* log n)，空间O(log n)

TreeNode * dfs(ListNode *head, int size){
	if(size <= 0) return NULL;
	if(size == 1) return new TreeNode(head->val);


	ListNode *cur = head;
	for(int i = 0; i < size/2; ++i){
		cur = cur->next;
	}
	TreeNode *root = new TreeNode(cur->val);
	root->left = dfs(head, size/2);
	root->right = dfs(cur->next, size - size/2 - 1 );
	return root;
}


TreeNode *sortedListToBST(ListNode *head) {
	ListNode *cur = head;
	int size = 0;
	while(cur){
		++size;
		cur = cur->next;
	}
	return dfs(head, size);
}

leetcode dfs Convert Sorted List to Binary Search Tree