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.

题目解析：

已知一个单链表，当中元素都是升序排列的，如今将链表转成平衡二叉树。

方法一：

我们用一个数组存储链表中元素，这样就能够利用下标訪问元素。之后依据二分查找法找树的每个节点就可以，代码例如以下：

/**
 * 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:
vector<int> list;
    TreeNode* makeTree(int start,int end)
    {
        if(start>end)
        return NULL;
        int mid=(start+end)/2;
        TreeNode* root=new TreeNode(list[mid]);
        TreeNode* left=makeTree(start,mid-1);
        TreeNode* right=makeTree(mid+1,end);
        root->left=left;
        root->right=right;
        return root;
    }
    TreeNode* sortedListToBST(ListNode* head) {
        if(head==NULL) return NULL;
        ListNode* p=head;
        while(p)
        {
            list.push_back(p->val);
            p=p->next;
        }
         return makeTree(0,list.size()-1);
    }
};

方法二

不用额外开辟空间，记录链表的长度以及表头就可以，代码例如以下：

class Solution {
public:
    int length(ListNode* head)
    {
        int len=0;
        ListNode* p=head;
        while(p)
        {
            len++;
            p=p->next;
        }
        return len;
    }
    TreeNode* makeTree(ListNode* head,int start,int end)
    {
        if(start>end)
        return NULL;
        int mid=(start+end)/2;
        ListNode* p=head;
        for(int i=start;i<mid;i++)
        {
            p=p->next;
        }
        TreeNode* root=new TreeNode(p->val);
        TreeNode* left=makeTree(head,start,mid-1);
        TreeNode* right=makeTree(p->next,mid+1,end);
        root->left=left;
        root->right=right;
        return root;
    }
    TreeNode* sortedListToBST(ListNode* head) {
        if(head==NULL) return NULL;
        ListNode* p=head;
       int len=length(p);
         return makeTree(p,0,len-1);
    }
};