leetcode109

1 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.
排序链表转换成平衡二叉搜索树
二叉搜索树就是左小右大，平衡树就是高度差小于等于1.可以想到根节点root一定是链表中间节点，左半部分又是一颗平衡二叉搜索树，则root->left就是左部分的中间节点，同时是左半部分的root，右边同理。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedListToBST(ListNode* head) {

        return ToBST(head,nullptr);
    }
    TreeNode* ToBST(ListNode* head, ListNode* end){
        if(head==end) return nullptr;
        if(head->next==end){
            TreeNode* root=new TreeNode(head->val);
            return root;
        }
        ListNode* mid=head;
        ListNode* temp=head;
        while(temp!=end&&temp->next!=end){
            mid=mid->next;
            temp=temp->next->next;
        }//中间节点
        TreeNode* root=new TreeNode(mid->val);
        root->left=ToBST(head,mid);
        root->right=ToBST(mid->next,end);
        return root;
    }
};