从给定的输入集实现 n 叉树

Question

I am trying to create a Tree from given inputs. One root will be there including child nodes and sub child nodes. I can implement the Tree where I can add a child node to a specific master node (where I already know the root). But, I am trying to figure out that what is the recommended way to implement the tree where we have to first find the root node from the given input set. As an, example:

There are n lines,
Value in each line represent the master node of that line number.
4 ( n = 4, next n lines)

1 ( 1 is a master node of 1, here line number is 1)
1 (1 is a master node of 2, here line number is 2)
2 (2 is a master node of 3, here line number is 3)
1 (1 is a master node of 4, here line number is 4)

So the tree should look like,

   1
 / | 
2  4
|
3

Here, we can see that root node is 1. But before knowing all the input values, we can not guess the root node. Before implementing the tree, should I first find out the root node from the inputs? Or, is there any another way?

Below code is to add a node into the tree:

class Node : 

    # Utility function to create a new tree node 
    def __init__(self ,key): 
        self.key = key 
        self.child = [] 

def printNodeLevelWise(root): 
    if root is None: 
        return

    queue = [] 
    queue.append(root) 

    while(len(queue) >0): 

        n = len(queue) 
        while(n > 0) : 

            # Dequeue an item from queue and print it 
            p = queue[0] 
            queue.pop(0) 
            print (p.key,) 

            # Enqueue all children of the dequeued item 
            for index, value in enumerate(p.child): 
                queue.append(value) 

            n -= 1
        print ("") 


root = Node(1) 
root.child.append(Node(2)) 
root.child.append(Node(4)) 
root.child[0].child.append(Node(3)) 
printNodeLevelWise(root) 

Above code will give implement this tree:

   1
 / | 
2  4
|
3

But, what is the recommended way to implement the same from the given input?

Answer 1

Assuming that:

• with "master" you mean "parent", and
• when a line has a self-reference -- such that i occurs on line i of the list of masters -- it is the root

...then you could proceed like this:

nodes = {} # dictionary of all nodes keyed by their value
root = None
n = int(input()) # get the number of nodes

for linenum in range(1, n+1):
    parentnum = int(input()) # get the "master" for this line
    # create the involved nodes that do not yet exist:
    if not linenum in nodes:
        nodes[linenum] = Node(linenum)
    if not parentnum in nodes:
        nodes[parentnum] = Node(parentnum)
    if parentnum == linenum: # a self-reference indicates the root
        root = nodes[parentnum]
    else: # set the parent-child relationship
        nodes[parentnum].child.append(nodes[linenum])