Search for a Range

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

Solution:
Usually, O(log n) is related to binary search.
binary search multiple times, worst O(N)

public class Solution {
    public int[] searchRange(int[] A, int target) {
        int[] result = {Integer.MAX_VALUE, Integer.MIN_VALUE};
        binarySearch(A, target, 0, A.length-1, result);
        
        if(result[0] == Integer.MAX_VALUE)
            result[0] = -1;
        if(result[1] == Integer.MIN_VALUE)
            result[1] = -1;
        
        return result;
    }
    
    public void binarySearch(int[] A, int target, int low, int high, int[] result){
        if(low > high) return;
        int mid = (low + high) / 2;
        if(target == A[mid]){ 
            result[0] = Math.min(result[0], mid);
            result[1] = Math.max(result[1], mid);
            binarySearch(A, target, low, mid-1, result);
            binarySearch(A, target, mid+1, high, result);
        }
        else if(target > A[mid]) 
            binarySearch(A, target, mid+1, high, result);
        else if(target < A[mid]) 
            binarySearch(A, target, low, mid-1, result);
    }
}