LeetCode 109. 有序链表转换为二叉树

题目描述链接：https://leetcode-cn.com/problems/convert-sorted-list-to-binary-search-tree/

解题思路：递归，分治。给出的链表为有序升序链表，找出中位数作为根节点，中位数之前的为左子树，之后为右子树。

如此递归在建立左子树和右子树即可。

LeetCode C++ 参考代码如下：

/**
 * 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 buildTree(head,NULL);
        
    }

    ListNode* getMid(ListNode* left,ListNode* right){
        ListNode *slow=left;
        ListNode *fast=left;
        while(fast!=right&&fast->next!=right){
            slow=slow->next;
            fast=fast->next;
            fast=fast->next;
        }
        return slow;
    }
    TreeNode* buildTree(ListNode* left,ListNode* right){
        if(left==right){
            return NULL;
        }
        ListNode* mid=getMid(left,right);
        TreeNode* root=new TreeNode(mid->val);
        root->left=buildTree(left,mid);
        root->right=buildTree(mid->next,right);
        return root;
    }
};