<prompt> Kutokana na idadi kamili safu nums ambapo vipengele ni sorted katika mzunguko wa kupanda, kubadilisha ni kwa urefu-balanced binary search mti. Andika kazi katika Java.  <response>: Here is the Java code to convert a sorted integer array into a height-balanced binary search tree:public TreeNode sortedArrayToBSTint nums ifnums null nums.length 0 return null; return convertToBSTnums, 0, nums.length - 1;private TreeNode convertToBSTint nums, int start, int end ifstart end return null; int mid start end - start 2; TreeNode root new TreeNodenumsmid; root.left convertToBSTnums, start, mid - 1; root.right convertToBSTnums, mid 1, end; return root;The sortedArrayToBST method checks for the base case where the array is empty or null. If one of these cases is true, then it returns null.Otherwise, it calls the convertToBST method, which recursively constructs a balanced binary search tree using a Divide and Conquer approach. We first find the middle element in our range start to end and make it the root. Then, we recursively build the left and right subtrees by calling convertToBST with the new ranges start to mid-1 and mid1 to end.At each call, the convertToBST method constructs a new node, sets its value to the middle element and assigns it left and right children returned by subsequent recursive calls. Finally, it returns the root node of the constructed BST. 