题目描写：
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
思路：采取类似折半查找的方式找到当前根节点，则当前根节点左侧的属于它的左子树部份，当前根节点右侧的属于它的右子树部份。再采取一样的方法，递归地对当前根节点的左右子树做相同的处理。
代码：
TreeNode * Solution::sortedArrayToBST(vector<int> &num)
{
if(num.size() == 0)
return NULL;
return sortedArrayToBSTCmp(num,0,num.size()⑴);
}
TreeNode * Solution::sortedArrayToBSTCmp(vector<int> &num,int head,int rear)
{
TreeNode * root = (TreeNode *)malloc(sizeof(TreeNode));
int mid = (head + rear) / 2;
root->val = num[mid];
if(mid + 1 <= rear)
root->right = sortedArrayToBSTCmp(num,mid+1,rear);
if(head <= mid - 1)
root->left = sortedArrayToBSTCmp(num,head,mid⑴);
return root;
}