排序列表轉換爲二分查找樹-LintCode

描述：

給出一個所有元素以升序排序的單鏈表，將它轉換成一棵高度平衡的二分查找樹

樣例：

               2
1->2->3  =>   / \
             1   3

思路：

折半取鏈表中的元素作爲其根節點，左半部分作爲其左子樹，右半部份作爲其右子樹。不斷遞歸。

相比於數組轉換二叉查找樹，多了些鏈表的遍歷而已。

AC代碼：

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @return: a tree node
     */
    TreeNode *sortedListToBST(ListNode *head) {
        // write your code here
        if(head==NULL)
            return NULL;
        int lenth=0;
        ListNode*pr=head;
        while(pr!=NULL)
        {
            lenth++;
            pr=pr->next;
        }
        if(lenth==1)
            return new TreeNode(head->val);
        else
        {
            lenth/=2;
            int i=0;
            pr=head;
            while(i<lenth-1)
            {
                pr=pr->next;
                i++;
            }
            TreeNode*root=new TreeNode(pr->next->val);
            root->right=sortedListToBST(pr->next->next);
            pr->next=NULL;
            root->left=sortedListToBST(head);
            return root;
        }
        
        
        
        
    }
};