def getFreeHoursInCommon1(a, b):
result = [[] for i in range(period)]
leader = [Node.inorderarr(a[x], [])
for x in range(period)] #without time in common
#leader = [[12, 13, 13.5], [10, 13.5, 15], [13.5, 15], [13.5, 14], [11, 13.5], [13.5], [13.5]] #with time in common
#print(leader)
for i in range(len(leader)):
for j in range(len(leader[i])):
if Node.search(b[i], leader[i][j]) != None and Node.search(
b[i], leader[i][j]) != b[i]:
result[i].append(leader[i][j])
return result
