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);
}
};