Answer:
d
Explanation:
The one action listed in the question that would suggest unfair labor practices would be if Hyper-Tech managers prohibited the distribution of union literature in the company cafeteria. This is because a company can prevent employees from partaking in other tasks and getting distracted while on duty. When an employee is on duty they are getting paid to focus and complete their responsibilities. An employer also has the right to prevent non-employee individuals from entering private property such as their facility. What a company cannot do is prevent their employees from deciding what the literature that they want to read or the decisions that they want to make outside of work hours. Therefore, preventing the employees from accepting literature while on their break time would be considered unfair labor practice.
Answer:
To change ur grade make sure to do it on the website first by right clicking your mouse and clicking inspect element and once done changing x out and it will save
Explanation:
Answer:
#
Explanation:
I have notes on it we learned it in 8th
Answer:
B.O(n).
Explanation:
Since the time complexity of visiting a node is O(1) in iterative implementation.So the time complexity of visiting every single node in binary tree is O(n).We can use level order traversal of a binary tree using a queue.Which can visit every node in O(n) time.Level order traversal do it in a single loop without doing any extra traversal.
Answer:
Explanation:
We start from the bottom-most and rightmost internal node of min Heap and then heapify all internal modes in the bottom-up way to build the Max heap.
To build a heap, the following algorithm is implemented for any input array.
BUILD-HEAP(A)
heapsize := size(A)
for i := floor(heapsize/2) downto 1
do HEAPIFY(A, i)
end for
Convert the given array of elements into an almost complete binary tree.
Ensure that the tree is a max heap.
Check that every non-leaf node contains a greater or equal value element than its child nodes.
If there exists any node that does not satisfy the ordering property of max heap, swap the elements.
Start checking from a non-leaf node with the highest index (bottom to top and right to left).
