LintCode 把排序数组转换为高度最小的二叉搜索树

题目描述：
给一个排序数组（从小到大），将其转换为一棵高度最小的排序二叉树。

注意事项

There may exist multiple valid solutions, return any of them.

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样例
给出数组 [1,2,3,4,5,6,7], 返回

4
/ \
2 6
/ \ / \
1 3 5 7

ac代码：

/** * 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 A: A sorted (increasing order) array     * @return: A tree node     */    TreeNode *dfs(vector<int> A,int l,int r)    {        if(l>r)            return NULL;        int mid=(r+l)/2;        TreeNode *root=new TreeNode(A[mid]);        root->left=dfs(A,l,mid-1);        root->right=dfs(A,mid+1,r);        return root;    }    TreeNode *sortedArrayToBST(vector<int> &A) {        // write your code here        if(A.size()==0)            return NULL;        int r,l,mid;        l=0,r=A.size()-1;        mid=(r+l)/2;        TreeNode *root=new TreeNode(A[mid]);        root->left=dfs(A,l,mid-1);        root->right=dfs(A,mid+1,r);        return root;    }};