109 - 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.

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思路分析：

这题的关键是能找出当前链表的中间节点，然后再递归左右的子链表，开始的时候程序先计算链表总厂，然后传入两个前后索引指针，最后每次递归找出中间节点即可。

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; *//** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int calLen(ListNode *node)    {        int len = 0;        while(node)        {            len++;            node = node->next;        }        return len;    }        TreeNode *createTree(ListNode *node, int left, int right)    {        if (left > right)            return NULL;                    int mid = (left + right) / 2;                ListNode *p = node;                for(int i = left; i < mid; i++)            p = p->next;                    TreeNode *leftNode = createTree(node, left, mid - 1);        TreeNode *rightNode = createTree(p->next, mid + 1, right);                TreeNode *tNode = new TreeNode(p->val);                tNode->left = leftNode;        tNode->right = rightNode;                return tNode;            }        TreeNode *sortedListToBST(ListNode *head) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int len = calLen(head);        return createTree(head, 0, len - 1);    }};