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Heaps: Python
Removing the Min

Min-heaps would be useless if we couldn’t retrieve the minimum value. We’ve gone through a lot of work to maintain that value because we’re going to need it!

Our goal is to efficiently remove the minimum element from the heap. You’ll recall that we always locate the minimum element at index 1, (a sentinel element occupies index 0).

Our internal list mirrors a binary tree. There’s a delicate balance of parent and child relationships we would ruin by directly removing the minimum.

print(heap.heap_list) # [None, 10, 21, 13, 61, 22, 23, 99] heap.retrieve_min() # 10 # [None, ???, 21, 13, 61, 22, 23, 99]

We need to remove an element that has no children; we need to remove the last element. Swap the minimum with the last element, and we can easily remove the minimum from the end of the list.

# [None, (10), 21, 13, 61, 22, 23, {99}] heap.retrieve_min() # [None, {99}, 21, 13, 61, 22, 23, (10)] # [None, 99, 21, 13, 61, 22, 23] # 10

Terrific! We removed the minimum element with minimal disruption. Unfortunately, our heap is out of shape again with 99 sitting where the minimum element should be.

Patience, young Heap-prentice. We will solve this in lessons to come…



Define a .retrieve_min() method within our MinHeap class. Its only parameter is self.

Check if our internal count is at 0

If it is, we have no elements to retrieve, so print a friendly “No items in heap” and return None.


Declare the variable min, which is the element at index 1 in our internal list.


Print the message “Removing: min from self.heap_list“.

Then, swap the element at index 1 with the last element in the internal list.

Remove the last element from the list, and decrement the count.


Print the message “Last element moved to first: self.heap_list“.

Finally, return the min variable.


Tab over to and run the test code.

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