*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

public TreeNode sortedArrayToBST(int[] num) {
if (num.length == 0)
return null;

return sortedArrayToBST(num, 0, num.length - 1);
}

public TreeNode sortedArrayToBST(int[] num, int start, int end) {
if (start > end)
return null;

int mid = (start + end) / 2;
TreeNode root = new TreeNode(num[mid]);
root.left = sortedArrayToBST(num, start, mid - 1);
root.right = sortedArrayToBST(num, mid + 1, end);

return root;
}

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