leetcode Convert Sorted List to Binary Search Tree

将单链表转换为平衡二叉查找树，因为单链表已经为升序排列，采用递归。

步骤

1，找到链表的中间节点，该节点作为当前的父节点

2，递归单链表中该父节点左部的所有节点以及右部的所有节点，中心节点分别作为该父节点的左指针和右指针。

注意，链表中只一个结点和两个结点时需要分情况讨论。

代码

class Solution
{
public:
	TreeNode *sortedListToBST(ListNode *head)
	{
		TreeNode *treeHead = NULL;
		if(head==NULL) 
			return treeHead;

		ListNode *end = head;

		while(end->next)
			end = end->next;

		return sortedListToBSTHelper(head, end);		
	
	}



	TreeNode *sortedListToBSTHelper(ListNode *start, ListNode *end)
	{

		TreeNode *treeNode = NULL;

		treeNode = new TreeNode(0);

		// 当前链表中只有一个结点的情况
		if(start==end) 
		{
			treeNode->val = start->val;
			return treeNode;
		}

		//当前链表中有两个结点的情况
		if(end == start->next)
		{
			treeNode->val = start->val;
			TreeNode *Node = new TreeNode(end->val);
			treeNode->right = Node;
			return treeNode;
		}
		
		ListNode *pre, *p, *prepre;
		pre = start;
		prepre = pre;
		p = start;

		//找到当前链表的中间结点
		while(p&&pre->next&&p->next&&p->next->next)
		{
			prepre = pre;
			pre = pre->next;
			p = p->next->next;		
		}
		
		treeNode->val = pre->val;
		prepre->next = NULL;

		//递归找到中间节点的左右结点
		treeNode->left= sortedListToBSTHelper(start, prepre);
		treeNode->right = sortedListToBSTHelper(pre->next, end);

		return treeNode;	
	
	}

};