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singly linked list feedback

I recently begun learning about data structures and algorithms. Here is a singly linked list code with all the functions I could come up with that I made for exercise. I am looking for ways to improve my code, I was thinking about using different methods (like findNode) inside others (like removeNode) but that's what I came up with. Any review, feedback and critique would be appreciated.
Code:
class Node(): def __init__(self, data, next=None): self.data = data self.next = next class LinkedList(): def __init__(self): self.head = None def build(self, array): # self.clear() # uncomment if self.build should clear the list before appending data to the linked list for data in array: self.append(data) def append(self, data): if self.head == None: self.head = Node(data) else: curr = self.head while curr.next != None: curr = curr.next curr.next = Node(data) def push(self, data): newNode = Node(data) newNode.next = self.head self.head = newNode def insert(self, index, data): if index == 0: self.push(Node(data)) return curr = self.head i = 0 while curr != None: if i == index - 1: # i == index - 1 because we are looking for the item before the index newNode = Node(data) newNode.next = curr.next curr.next = newNode curr = curr.next i += 1 if index > i: print('linked list index out of range!') def remove(self, data): if self.empty(): print('empty linked list!') return if self.head.data == data: self.head = self.head.next return curr = self.head while curr != None: if curr.next.data == data: curr.next = curr.next.next break curr = curr.next else: print('no node was found to remove!') def removeAtIndex(self, index): if self.empty(): print('empty linked list!') return if index == 0: self.head = self.head.next return curr = self.head i = 0 while curr.next != None: if i == index - 1: curr.next = curr.next.next break i += 1 curr = curr.next else: print('no node was found to remove!') def find(self, index): if self.empty(): print('empty linked list!') return curr = self.head i = 0 while curr != None: if i == index: return curr.data i += 1 curr = curr.next return 'no node at this index was found!' def findIndex(self, data): if self.empty(): print('empty linked list!') return curr = self.head i = 0 while curr != None: if curr.data == data: return i i += 1 curr = curr.next return 'no node was found!' def replace(self, nodeData, data): if self.empty(): print('empty linked list!') return if nodeData == self.head.data: newNode = Node(data) newNode.next = self.head self.head = newNode return curr = self.head while curr != None: if curr.next.data == nodeData: newNode = Node(data) newNode.next = curr.next.next curr.next = newNode break curr = curr.next else: print('no node was found!') def replaceAtIndex(self, index, data): if self.empty(): print('empty linked list!') return if index == 0: newNode = Node(data) newNode.next = self.head.next self.head = newNode return i = 0 curr = self.head while curr != None: if i == index - 1: newNode = Node(data) newNode.next = curr.next.next curr.next = newNode break i += 1 curr = curr.next else: print('linked list index out of range!') def count(self): curr = self.head i = 0 while curr != None: i += 1 curr = curr.next return f'Number of nodes: {i}' def show(self): nodes = [] curr = self.head while curr != None: nodes.append(curr.data) curr = curr.next return nodes def clear(self): self.head = None def empty(self): if self.head == None: return True return False # ll = LinkedList() # ll.append(5) # ll.build([2, 7, 0, 11]) # print(ll.show()) # ll.push(0) # ll.remove(0) # print(ll.show()) # ll.append(26) # ll.removeAtIndex(0) # print(ll.show()) # ll.insert(6, 15) # ll.insert(5, 15) # print(ll.show()) # ll.replaceAtIndex(6, 24) # ll.replace(24, 30) # print(ll.show()) 
There are also some tests in the commens at the bottom.
submitted by ViktorCodes to learnpython

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