K Closest In Sorted Array in Python(Binary Search)

Given a target integer T, a non-negative integer K and an integer array A sorted in ascending order, find the K closest numbers to T in A.

Assumptions:

A is not null
K is guranteed to be >= 0 and K is guranteed to be <= A.length

Return:

A size K integer array containing the K closest numbers(not indices) in A, sorted in ascending order by the difference between the number and T.

Examples:

A = {1, 2, 3}, T = 2, K = 3, return {2, 1, 3} or {2, 3, 1}
A = {1, 4, 6, 8}, T = 3, K = 3, return {4, 1, 6}

class Solution(object):
  def kClosest(self, array, target, k):
    A = array
    if not A or k ==0:
      return []
    res = []
    index = self.binarySearch(A,target)
    res.append(A[index])
    k -= 1
    left = index - 1
    right = index + 1
    while k > 0:
      if left >= 0 and right <= len(A) - 1:
        if abs(A[left] - target) <= abs(A[right] - target):
          res.append(A[left])
          left -= 1
          k -= 1
        else:
          res.append(A[right])
          right += 1
          k -= 1
      if left < 0:
        while k > 0 and right <= len(A) - 1:
          res.append(A[right])
          right += 1
          k -= 1
        break
      if right > len(A) - 1:
        while k > 0 and left >= 0:
          res.append(A[left])
          left -= 1
          k -= 1
        break
    return res
  def binarySearch(self,L,target):
    low = 0
    high = len(L) - 1
    while low < high - 1:
      mid = (low + high)/2
      if L[mid] == target:
        return mid
      elif L[mid] > target:
        high = mid
      else: low = mid
    return low if abs(L[low] - target) <= abs(L[high] - target) else high