Given a set of parent nodes where the index of the array is the child of each Node value, the task is to insert the nodes as a forest(multiple trees combined together) where each parent could have more than two children. After inserting the nodes, print each level in a sorted format.
Example:
Input: arr[] = {5, 3, 1, 2, 5, 3}
Output:
1
2
3
1 5Input: arr[] = {1, 1, 1, 1, 1, 1}
Output:
1
0 1 2 3 4
5
Below is the explanation of the above examples:

Example 1:
 In this given array, the elements of the array will be the parent node and the array index will be the child nodes.
 Initially, we set the root of the forest to be 1 for reference.
 Now on traversing the array, we insert the nodes into the forest structure.
 Initially we identify the roots of the individual trees in the forest and insert them into the root of the forest.
 The index of 1 is 2. Print 1 and append 2 as child node.
 Now search the list for list value as 2. Index 3 has value 2. Therefore 3 becomes the child of 2.
 Now the indexes having value 3 are 1 and 5. So 1 and 5 are the children of 3.
 The list does not contain 1 so ignore 1.
 The index that contains 5 are 0 and 4. So they become the child.
1  root of the forest / 2  level (0) / 3  level (1) / \ 1 5  level (2) / \ 0 4  level (3) Note: level (0) contains roots of each tree

Example 2:
 In this case, the tree will be of the format
1  root of the forest /    \ 0 1 2 3 4  level (0)  5  level (1) Note: level (0) contains roots of each tree
Prerequisite: Level order traversal.
Approach: The idea is to recursively insert nodes in a tree. However the tree structure is quite different, usually in the case of binary tree there will be a maximum of two child nodes for any node but in this case the root node can have N number of child nodes.’1′ is considered as the root and the index of the root will be considered as child nodes.
Example:
If 1 is present in index 3 then 3 will be the child node of 1.
1 / 3
Insert 1 into the queue. Now if the root is empty then 1 node becomes the root. Now dequeue and queue the child nodes of 1. Create nodes and append them with the root. Continue this till all the child nodes have been inserted.
Level order Traversal: 1 3 5 2 4 6 9 The output for level order traversal will be: 1 3 5 2 4 6 9
Same enqueue and dequeue approach is followed for traversing by level order.
Below is the implementation of the above approach:
# Python3 implementation of the approach # Node creation class Node:
# Constructor
def __init__( self , data):
self .val = data
# Since n children are possible for a root.
# A list created to store all the children.
self .child = []
# Function to insert def insert(root, parent, node):
# Root is empty then the node will become the root
if root is None :
root = node
else :
if root.val = = parent:
root.child.append(node)
else :
# Recursive approach to
# insert the child
l = len (root.child)
for i in range (l):
if root.child[i].val = = parent:
insert(root.child[i], parent, node)
else :
insert(root.child[i], parent, node)
# Function that calls levelorder method to # perform level order traversal def levelorder_root(root):
if root:
level = []
level.append(root)
print (root.val)
levelorder(level)
# Function to perform level order traversal def levelorder(prev_level):
cur_level = []
print_data = []
l = len (prev_level)
if l = = 0 :
exit()
for i in range (l):
prev_level_len = len (prev_level[i].child)
for j in range (prev_level_len):
# enqueue all the children
# into cur_level list
cur_level.append(
prev_level[i].child[j])
# Copies the entire cur_level
# list into prev_level
print_data.append(
prev_level[i].child[j].val)
prev_level = cur_level[:]
print ( * print_data)
levelorder(prev_level)
# Driver code # 1 is the root element arr = [  1 ,  1 ,  1 ,  1 ,  1 ]
root = Node(  1 )
l = len (arr)
que = []
# Inserting root element to the queue que.append(  1 )
while 1 :
temp = []
for i in range (l):
if arr[i] in que:
# Insert elements into the tree
insert(root, arr[i], Node(i))
temp.append(i)
# Append child nodes into the queue
# and insert the child
que = temp[:]
if len (que) = = 0 :
break
levelorder_root(root) 
1 0 1 2 3 4
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