*Convert Sorted Array to Binary Search Tree

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

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

 1     public TreeNode sortedArrayToBST(int[] num) {
 2         if (num.length == 0)
 3             return null;
 4  
 5         return sortedArrayToBST(num, 0, num.length - 1);
 6     }
 7  
 8     public TreeNode sortedArrayToBST(int[] num, int start, int end) {
 9         if (start > end)
10             return null;
11  
12         int mid = (start + end) / 2;
13         TreeNode root = new TreeNode(num[mid]);
14         root.left = sortedArrayToBST(num, start, mid - 1);
15         root.right = sortedArrayToBST(num, mid + 1, end);
16  
17         return root;
18     }

reference: http://www.programcreek.com/2013/01/leetcode-convert-sorted-array-to-binary-search-tree-java/

http://articles.leetcode.com/2010/11/convert-sorted-array-into-balanced.html