How to convert a list of edges to a tree in python?

I am having a list of edges that has the following format:

edges=[[1,4],[1,3],[1,2],[3,5],[3,6],[3,7]]

Here in each edge the first element is the parent node and the second is a child node i.e, in

[1,4]---->(1 is the parent node and 4 is child node)

I have to create a function that returns the pointer to the root of the tree. At first I tried creating a dictionary but after creating I am unable to proceed.

Please provide any ideas of how to implement this?

There are many ways to create a tree data structure... moreover Python does not have a pointer data type, so the root of a tree would be an object.

Here is one way to do it:

First define a Node class:

class Node():
    def __init__(self, data=None):
        self.data = data
        self.children = []

Then the main algorithm:

def create_tree(edges):
    # Get all the unique keys into a set
    node_keys = set(key for keys in edges for key in keys)
    # Create a Node instance for each of them, keyed by their key in a dict:
    nodes = { key: Node(key) for key in node_keys }
    # Populate the children attributes from the edges
    for parent, child in edges:
        nodes[parent].children.append(nodes[child])
        # Remove the child from the set, so we will be left over with the root
        node_keys.remove(child)
    # Get the root from the set, which at this point should only have one member
    for root_key in node_keys:  # Just need one
        return nodes[root_key]

Run it as follows:

# Example run
edges = [[1,4],[1,3],[1,2],[3,5],[3,6],[3,7]]
root = create_tree(edges)

If you want to quickly verify the tree's shape, add this method to the Node class:

    def __repr__(self, indent=""):
        return (indent + repr(self.data) + "\n"
                + "".join(child.__repr__(indent+"  ") 
                          for child in self.children))

And use it to print the tree:

print(root)

This output string is just a very simple way to visualise the tree. With a bit more code you can also draw the branches of the tree, but this is enough to debug the code.