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Program to find the root of a n-ary tree in Python
Suppose, we are given the nodes of an n-ary tree in an array. We have to find and return the root node of the tree by reconstructing it. The full tree has to be displayed from the returned node in preorder notation.
So, if the input is like
then the output will be
[14, 27, 32, 42, 56, 65]
We will use the root of the tree to display the pre order traversal of the tree. So, the output is a pre order traversal of the tree.
To solve this, we will follow these steps −
indegree := a new map containing integer values
for each node in tree, do
for each child in children pointers of the node, do
indegree[value of child] := indegree[value of child] + 1
for each node in tree, do
if indegree[value of node] is same as 0, then
return node
return null
Example (Python)
Let us see the following implementation to get better understanding −
import collections
class Node:
def __init__(self, value, child = None) -> None:
self.val = value
self.children = []
if child != None:
for value in child:
self.children.append(value)
def solve(tree):
indegree = collections.defaultdict(int)
for node in tree:
for child in node.children:
indegree[child.val] += 1
for node in tree:
if indegree[node.val] == 0:
return node
return None
def treeprint(node, tree):
if node == None:
tree.append("None")
return tree
if tree == None:
tree = []
tree.append(node.val)
for child in node.children:
treeprint(child, tree)
return tree
node6 = Node(65)
node5 = Node(56)
node4 = Node(42, [node5, node6])
node3 = Node(32)
node2 = Node(27)
node1 = Node(14, [node2, node3, node4])
tree = [node2, node1, node5, node3, node6, node4]
root = solve(tree)
print(treeprint(root, None))Input
node6 = Node(65) node5 = Node(56) node4 = Node(42, [node5, node6]) node3 = Node(32) node2 = Node(27) node1 = Node(14, [node2, node3, node4]) tree = [node2, node1, node5, node3, node6, node4]
Output
[14, 27, 32, 42, 56, 65]
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