convert a sorted array into a balanced binary search tree

Question:
Given a sorted array, and converted it into a balanced binary search tree. 
Solution: If you would have to choose an array element to be the root of a balanced BST, which element would you pick? The root of a balanced BST should be the middle element from the sorted array. You would pick the middle element from the sorted array in each iteration. You then create a node in the tree initialized with this element. After the element is chosen, what is left? Could you identify the sub-problems within the problem? There are two arrays left — The one on its left and the one on its right. These two arrays are the sub-problems of the original problem, since both of them are sorted. Furthermore, they are subtrees of the current node’s left and right child.
The code below creates a balanced BST from the sorted array in O(N) time (N is the number of elements in the array). Compare how similar the code is to a binary search algorithm. Both are using the divide and conquer methodology.

Node sortedArrayToBST(int arr[], int start, int end) {
	if (start > end) return null;
	// same as (start+end)/2, avoids overflow.
	int mid = start + (end - start) / 2;
	Node node = new Node(arr[mid]);
	node.leftChild = sortedArrayToBST(arr, start, mid-1);
	node.rightChild = sortedArrayToBST(arr, mid+1, end);
	return node;
}	 
Node sortedArrayToBST(int arr[], int n) {
	return sortedArrayToBST(arr, 0, n-1);
}

From:  http://www.ihas1337code.com/2010/11/convert-sorted-array-into-balanced.html
확장: 이 문제는 종종 다른 문제의 하위 문제입니다. 예를 들어 어떻게 두 개의 BST를 하나의 BST로 합치는지, 시간의 복잡도를 O(N)로 요구합니다.대체적인 사고방식은 in-order walk로 두 그룹의 정렬된 수조(O(N)를 얻고 두 개의 수조(O(N)를 합친 다음에 수조를 균형 두 갈래 나무(O(N)로 바꾸는 것이다.