LeetCode OJ——Convert Sorted List to Binary Search Tree

http://oj.leetcode.com/problems/convert-sorted-list-to-binary-search-tree/

将一个按照元素升序排列的链表转换成BST。根据自身性质“元素升序排列”，再参考将升序排列的数组转换成BST的上一道题目，只需要将对数组的处理方式变成对链表的。还是得好好利用元素已经升序排列这个性质，不用那种纯粹的重新生成一个BST，按照左转、右转、先左转后右转、先右转后左转。

#include <iostream>
#include <vector>

using namespace std;

struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};

struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
void fun(ListNode *begin ,int len,TreeNode *node)
{
if(len == 1)
{
node->val = begin->val;
return;
}
int leftlen = len/2;
int rightlen = len - len/2 -1;

ListNode *midNode = begin;
int tt = leftlen;
while(tt--)
midNode = midNode->next;
node->val = midNode->val;

if(leftlen)
{
TreeNode *nodeLeft = new TreeNode(0);
node->left = nodeLeft;
fun(begin,leftlen,nodeLeft);
}
if(rightlen)
{
TreeNode *nodeRight = new TreeNode(0);
node->right = nodeRight;
fun(midNode->next,rightlen,nodeRight);
}
}

TreeNode *sortedListToBST(ListNode *head) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
if(head == NULL)
return NULL;

TreeNode *root = new TreeNode(0);
//length
ListNode *temp= head;
int len = 0;
while(temp)
{
len++;
temp = temp->next;
}
fun(head,len,root);

return root;
}
};

int main()
{
Solution *mySolution = new Solution();

ListNode *head = new ListNode(1);
ListNode *n1 = new ListNode(3);
head->next = n1;
ListNode *n2 = new ListNode(5);
n1->next = n2;
ListNode *n3 = new ListNode(7);
n2->next = n3;
ListNode *n4 = new ListNode(9);
n3->next = n4;
ListNode *n5 = new ListNode(11);
n4->next = n5;
mySolution->sortedArrayToBST(head);

return 0;
}