leetCode # Convert Sorted List to Binary Search Tree

题目：将排好序的链表转化为二叉搜索树

分析：算法思路是取中间为根节点，左边是左子树，右边是右子树，然后递归

答案：

/** * 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 {private:TreeNode* createTree(ListNode* head, int left, int right){    if (left > right)        return NULL;    int mid = (left+right)/2;    // find mid node    ListNode* cNode = head;    for (int i = left; i < mid; i++){        cNode = cNode->next;    }    TreeNode* tNode = new TreeNode(cNode->val);    tNode->left = createTree(head, left,mid-1);    tNode->right = createTree(cNode->next, mid+1, right);    return tNode;}int getLength(ListNode* head){    int len = 0;    ListNode* p = head;    while(p){        p = p->next;        len++;    }    return len;}public:    TreeNode* sortedListToBST(ListNode* head) {        createTree(head,0,getLength(head)-1);    }};

注意：很有意思的一个问题是边界的判断。我一开始把最后一句写成createTree(head,0,getLength(head))就出现了无限循环