LeetCode109. Convert Sorted List to Binary Search Tree

题目：

https://leetcode.com/problems/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.

思路：

题意：给一个排好序的链表，把它转化成高度平衡的二叉查找树。
其实就是不断的选出中间的节点作为根结点，分成两段，再分别往下递归选出中间节点。

代码：

/**
 * 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* partition(ListNode* head,int len){
        if(head==NULL || len<=0){
            return NULL;
        }
        ListNode* left=head;
        ListNode* right=head;
        int partLen=len/2;
        for(int i=1;i<=partLen;i++){
            right=right->next;
        }
        TreeNode* treeNode=new TreeNode(right->val);
        right=right->next;
        TreeNode* leftChild=partition(left,partLen);
        TreeNode* rightChild=partition(right,len-partLen-1);
        treeNode->left=leftChild;
        treeNode->right=rightChild;
        return treeNode;
    }
    TreeNode* sortedListToBST(ListNode* head) {
        int len=0;
        if(head==NULL){
            return NULL;
        }
        ListNode* p=head;
        while(p!=NULL){
            len++;
            p=p->next;
        }
        return partition(head,len);
    }
};