find the smallest node in a linked list in python

To find the smallest node in a linked list in Python, you can traverse the linked list and keep track of the smallest node encountered so far. Here's a sample implementation:

main.py
class Node:
    def __init__(self, data=None):
        self.data = data
        self.next = None

def find_smallest_node(head):
    if head is None:
        # Empty linked list
        return None

    smallest_node = head
    current_node = head.next

    while current_node is not None:
        if current_node.data < smallest_node.data:
            smallest_node = current_node
        current_node = current_node.next

    return smallest_node
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In this implementation, we start by checking if the linked list is empty (i.e. head is None). If it is, we simply return None.

Otherwise, we initialize smallest_node to be the first node in the linked list, and current_node to be the second node. We then traverse the linked list using a while loop until we reach the end of the list (i.e. current_node is None).

During each iteration, we compare current_node's data to smallest_node's data. If current_node's data is smaller, we update smallest_node. We then move on to the next node in the linked list by setting current_node to its next node.

Once we've reached the end of the linked list, we return smallest_node.

You can use this function as follows:

main.py
# Create a linked list
head = Node(3)
node2 = Node(5)
node3 = Node(1)
node4 = Node(4)
head.next = node2
node2.next = node3
node3.next = node4

# Find the smallest node
smallest_node = find_smallest_node(head)
print(smallest_node.data)  # Output: 1
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