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Construct a Complete N-ary Tree from given Postorder Traversal

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  • Difficulty Level : Hard
  • Last Updated : 25 Jul, 2022
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Given an array arr[] of size M that contains the post-order traversal of a complete N-ary tree, the task is to generate the N-ary tree and print its preorder traversal.

A complete tree is a tree where all the levels of the tree are completely filled, except may be for the last level but the nodes in the last level are as left as possible.

Examples:

Input: arr[] = {5, 6, 7, 2, 8, 9, 3, 4, 1}, N = 3
Output: 

Example-1

The tree structure for above example

Input: arr[] = {7, 8, 9, 2, 3, 4, 5, 6, 1}, N = 5
Output: 

Example 2

Complete structure for the 2nd example

 

Approach: The approach to the problem is based on the following facts about the complete N-ary tree:

Say any subtree of the complete N-ary tree has a height H. So there are total H+1 levels numbered from 0 to H.

Total number of nodes in first H levels are N0 + N1 + N2 + . . . + NH-1 = (NH – 1)/(N – 1)

The maximum possible nodes in the last level is NH
So if the last level of the subtree has at least NH nodes, then total nodes in the last level of the subtree is  (NH – 1)/(N – 1) +  NH 

The height H can be calculated as ceil[logN( M*(N-1) +1)] – 1 because 
N-ary complete binary tree there can have at max (NH+1 – 1)/(N – 1)] nodes

Since the given array contains the post-order traversal, the last element of the array will always be the root node. Now based on the above observation the remaining array can be partitioned into a number of subtrees of that node.

Follow the steps mentioned below to solve the problem:

  1. The last element of the array is the root of the tree.
  2. Now break the remaining array into subarrays which represent the total number of nodes in each subtree.
  3. Each of these subtrees surely has a height of H-2 and based on the above observation if any subtree has more than (NH-1 – 1)/(N – 1) nodes then it has a height of H-1.
  4. To calculate the number of nodes in subtrees follow the below cases: 
    • If the last level has more than NH-1 nodes, then all levels in this subtree are full and the subtree has (NH-1-1)/(N-1) + NH-1 nodes.
    • Otherwise, the subtree will have (NH-1-1)/(N-1) + L nodes where L is the number of nodes in the last level.
    • L can be calculated as L = M – (NH – 1)/(N – 1)
  5. To generate each subarray repeatedly apply the 2nd to 4th steps adjusting the size (M) and height (H) accordingly for each subtree.
  6. Return the preorder traversal of the tree formed in this way

Follow the illustration below for a better understanding

Illustration:

Consider the example: arr[] = {5, 6, 7, 2, 8, 9, 3, 4, 1}, N = 3.

So height (H) = ceil[ log3(9*2 + 1) ] – 1 = ceil[log319] – 1 = 3 – 1 = 2 and L = 9 – (32 – 1)/2 = 5.

1st step: So the root = 1

1st step of forming the tree

1st step of forming the tree

2nd step: The remaining array will be broken into subtrees. For the first subtree H = 2.
5 > 32-1 i.e., L > 3. So the last level is completely filled and the number of nodes in the first subtree = (3-1)/2 + 3 = 4 [calculated using the formulae shown above].

The first subtree contains element = {5, 6, 7, 2}. The remaining part is {8, 9, 3, 4}
The root of the subtree = 2.

Now when the subtree for 5, 6, and 7 are calculated using the same method they don’t have any children and are the leaf nodes themselves.

2nd step of generating the tree

2nd step of generating the tree

3rd step: Now L is updated to 2. 
2 < 3. So using the above formula, number of nodes in the second subtree is 1 + 2 = 3. 

So the second subtree have elements {8, 9, 3} and the remaining part is {4} and 3 is the root of the second subtree.

3nd step of generating the tree

3nd step of generating the tree

4th step: L = 0 now and the only element of the subtree is {4} itself. 

final structure of tree

Final structure of tree

After this step the tree is completely built.

Below is the implementation of the above approach:

C++




// C++ code to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Node Class
template <typename T> class Node {
public:
    Node(T data);
 
    // Get the first child of the Node
    Node* get_first_child() const;
 
    // Get the Next Sibling of The node
    Node* get_next_sibling() const;
 
    // Sets the next sibling of the node
    void set_next_sibling(Node* node);
 
    // Sets the next child of the node
    void set_next_child(Node* node);
 
    // Returns the data the node contains
    T get_data();
 
private:
    T data;
 
    // We use the first child/next sibling representation
    // to represent the Tree
    Node* first_child;
    Node* next_sibling;
};
 
// Using template for generic usage
template <typename T> Node<T>::Node(T data)
{
    first_child = NULL;
    next_sibling = NULL;
    this->data = data;
}
 
// Function to get the first child
template <typename T>
Node<T>* Node<T>::get_first_child() const
{
    return first_child;
}
 
// Function to get the siblings
template <typename T>
Node<T>* Node<T>::get_next_sibling() const
{
    return next_sibling;
}
 
// Function to set next sibling
template <typename T>
void Node<T>::set_next_sibling(Node<T>* node)
{
    if (next_sibling == NULL) {
        next_sibling = node;
    }
    else {
        next_sibling->set_next_sibling(node);
    }
}
 
// Function to get the data
template <typename T> T Node<T>::get_data() { return data; }
 
// Function to set the child node value
template <typename T>
void Node<T>::set_next_child(Node<T>* node)
{
    if (first_child == NULL) {
        first_child = node;
    }
    else {
        first_child->set_next_sibling(node);
    }
}
 
// Function to construct the tree
template <typename T>
Node<T>* Construct(T* post_order_arr, int size, int k)
{
    Node<T>* Root = new Node<T>(post_order_arr[size - 1]);
    if (size == 1) {
        return Root;
    }
    int height_of_tree
        = ceil(log2(size * (k - 1) + 1) / log2(k)) - 1;
    int nodes_in_last_level
        = size - ((pow(k, height_of_tree) - 1) / (k - 1));
    int tracker = 0;
    while (tracker != (size - 1)) {
        int last_level_nodes
            = (pow(k, height_of_tree - 1)
               > nodes_in_last_level)
                  ? nodes_in_last_level
                  : (pow(k, height_of_tree - 1));
        int nodes_in_next_subtree
            = ((pow(k, height_of_tree - 1) - 1) / (k - 1))
              + last_level_nodes;
 
        Root->set_next_child(
            Construct(post_order_arr + tracker,
                      nodes_in_next_subtree, k));
 
        tracker = tracker + nodes_in_next_subtree;
        nodes_in_last_level
            = nodes_in_last_level - last_level_nodes;
    }
    return Root;
}
 
// Function to print the preorder traversal
template <typename T> void preorder(Node<T>* Root)
{
    if (Root == NULL) {
        return;
    }
    cout << Root->get_data() << "  ";
    preorder(Root->get_first_child());
    preorder(Root->get_next_sibling());
}
 
// Driver code
int main()
{
    int M = 9, N = 3;
    int arr[] = { 5, 6, 7, 2, 8, 9, 3, 4, 1 };
   
    // Function call
    preorder(Construct(arr, M, N));
    return 0;
}

Java




// Java code to implement the approach
 
import java.io.*;
import java.util.*;
 
class GFG
{
 
  // Function to calculate the
  // log base 2 of an integer
  public static int log2(int N)
  {
 
    // calculate log2 N indirectly
    // using log() method
    int result = (int)(Math.log(N) / Math.log(2));
 
    return result;
  }
 
  // Node Class
  public static class Node {
    int data;
 
    // We use the first child/next sibling
    // representation to represent the Tree
    Node first_child;
    Node next_sibling;
 
    // Initiallizing Node using Constructor
    public Node(int data)
    {
      this.first_child = null;
      this.next_sibling = null;
      this.data = data;
    }
 
    // Function to get the first child
    public Node get_first_child()
    {
      return this.first_child;
    }
 
    // Function to get the siblings
    public Node get_next_sibling()
    {
      return this.next_sibling;
    }
 
    // Function to set the next sibling
    public void set_next_sibling(Node node)
    {
      if (this.next_sibling == null) {
        this.next_sibling = node;
      }
      else {
        this.next_sibling.set_next_sibling(node);
      }
    }
 
    // Function to get the data
    public int get_data() { return this.data; }
 
    // Function to set the child node values
    public void set_next_child(Node node)
    {
      if (this.first_child == null) {
        this.first_child = node;
      }
      else {
        this.first_child.set_next_sibling(node);
      }
    }
  }
 
  public static void main(String[] args)
  {
    int M = 9, N = 3;
    int[] arr = { 5, 6, 7, 2, 8, 9, 3, 4, 1 };
 
    // Function call
    preorder(Construct(arr, 0, M, N));
  }
 
  // Function to print the preorder traversal
  public static void preorder(Node Root)
  {
    if (Root == null) {
      return;
    }
    System.out.println(Root.get_data() + "  ");
    preorder(Root.get_first_child());
    preorder(Root.get_next_sibling());
  }
 
  // Function to construct the tree
  public static Node Construct(int[] post_order_arr,
                               int tracker, int size,
                               int k)
  {
    Node Root = new Node(post_order_arr[tracker+size - 1]);
    if (size == 1) {
      return Root;
    }
    int height_of_tree
      = (int)Math.ceil(log2(size * (k - 1) + 1)
                       / log2(k))
      - 1;
    int nodes_in_last_level
      = size
      - (((int)Math.pow(k, height_of_tree) - 1)
         / (k - 1));
    int x=tracker;
    while (tracker != (size - 1)) {
      int last_level_nodes
        = ((int)Math.pow(k, height_of_tree - 1)
           > nodes_in_last_level)
        ? nodes_in_last_level
        : ((int)Math.pow(k,
                         height_of_tree - 1));
      int nodes_in_next_subtree
        = (((int)Math.pow(k, height_of_tree - 1)
            - 1)
           / (k - 1))
        + last_level_nodes;
 
      Root.set_next_child(Construct(
        post_order_arr, tracker, nodes_in_next_subtree, k));
 
      tracker = tracker + nodes_in_next_subtree;
      nodes_in_last_level
        = nodes_in_last_level - last_level_nodes;
    }
    return Root;
  }
}

Python3




# Python code to implement the approach
import math
 
# Node Class
class Node:
    # We use the first child/next sibling representation to represent the Tree
    def __init__(self, data):
        self.data = data
        self.first_child = None
        self.next_sibling = None
 
    # Function to get the first child
    def get_first_child(self):
        return self.first_child
 
    # Function to get the siblings
    def get_next_sibling(self):
        return self.next_sibling
 
    # Function to set next sibling
    def set_next_sibling(self, node):
        if self.next_sibling is None:
            self.next_sibling = node
        else:
            self.next_sibling.set_next_sibling(node)
 
    # Function to set the child node value
    def set_next_child(self, node):
        if self.first_child is None:
            self.first_child = node
        else:
            self.first_child.set_next_sibling(node)
 
    # Function to get the data
    def get_data(self):
        return self.data
 
# Function to construct the tree
def Construct(post_order_arr, size, k):
 
    Root = Node(post_order_arr[size - 1])
    if size == 1:
        return Root
    height_of_tree = math.ceil(
        math.log(size * (k - 1) + 1, 2) / math.log(k, 2)) - 1
 
    nodes_in_last_level = size - ((pow(k, height_of_tree) - 1) / (k - 1))
    tracker = 0
 
    while tracker != (size - 1):
        if pow(k, height_of_tree-1) > nodes_in_last_level:
            last_level_nodes = int(nodes_in_last_level)
        else:
            last_level_nodes = int((pow(k, height_of_tree - 1)))
 
        nodes_in_next_subtree = int(
            ((pow(k, height_of_tree - 1) - 1) / (k - 1)) + last_level_nodes)
 
        Root.set_next_child(
            Construct(post_order_arr[tracker:], nodes_in_next_subtree, k))
 
        tracker = tracker + nodes_in_next_subtree
        nodes_in_last_level = nodes_in_last_level - last_level_nodes
 
    return Root
 
# Function to print the preorder traversal
def preorder(root):
    if root == None:
        return
    print(root.get_data(), end=" ")
    preorder(root.get_first_child())
    preorder(root.get_next_sibling())
 
 
# Driver code
if __name__ == '__main__':
    M = 9
    N = 3
    arr = [5, 6, 7, 2, 8, 9, 3, 4, 1]
 
    # Function call
    preorder(Construct(arr, M, N))
 
# This code is contributed by Tapesh(tapeshdua420)

C#




// C# code to implement the approach
using System;
using System.Collections.Generic;
using System.Linq;
 
// Node Class
class Program {
 
    public class Node {
        int data;
 
        // We use the first child/next sibling
        // representation to represent the Tree
        Node first_child;
        Node next_sibling;
 
        public Node(int data)
        {
            this.first_child = null;
            this.next_sibling = null;
            this.data = data;
        }
 
        // Function to get the first child
        public Node get_first_child()
        {
            return this.first_child;
        }
 
        // Function to get the siblings
        public Node get_next_sibling()
        {
            return this.next_sibling;
        }
 
        // Function to set next sibling
        public void set_next_sibling(Node node)
        {
            if (this.next_sibling == null) {
                this.next_sibling = node;
            }
            else {
                this.next_sibling.set_next_sibling(node);
            }
        }
 
        // Function to get the data
        public int get_data() { return this.data; }
 
        // Function to set the child node value
        public void set_next_child(Node node)
        {
            if (this.first_child == null) {
                this.first_child = node;
            }
            else {
                this.first_child.set_next_sibling(node);
            }
        }
    }
    // Driver code
    public static void Main(string[] args)
    {
        int M = 9, N = 3;
        int[] arr = { 5, 6, 7, 2, 8, 9, 3, 4, 1 };
 
        preorder(Construct(arr, M, N));
    }
    // Function to print the preorder traversal
    public static void preorder(Node Root)
    {
        if (Root == null) {
            return;
        }
        Console.Write(Root.get_data() + "  ");
        preorder(Root.get_first_child());
        preorder(Root.get_next_sibling());
    }
 
    // Function to construct the tree
    public static Node Construct(int[] post_order_arr,
                                 int size, int k)
    {
        Node Root = new Node(post_order_arr[size - 1]);
        if (size == 1) {
            return Root;
        }
 
        int height_of_tree = (int)Math.Ceiling(
            (double)(Math.Log(size * (k - 1) + 1, 2)
                     / Math.Log(k, 2))
            - 1);
        // Console.WriteLine(height_of_tree);
 
        int nodes_in_last_level
            = size
              - (((int)Math.Pow(k, height_of_tree) - 1)
                 / (k - 1));
        int tracker = 0;
        while (tracker != (size - 1)) {
            int last_level_nodes
                = ((int)Math.Pow(k, height_of_tree - 1)
                   > nodes_in_last_level)
                      ? nodes_in_last_level
                      : ((int)Math.Pow(k,
                                       height_of_tree - 1));
 
            int nodes_in_next_subtree
                = (int)((Math.Pow(k, height_of_tree - 1)
                         - 1)
                        / (k - 1))
                  + last_level_nodes;
 
            Root.set_next_child(
                Construct((post_order_arr.Skip(tracker))
                              .Cast<int>()
                              .ToArray(),
                          nodes_in_next_subtree, k));
 
            tracker = tracker + nodes_in_next_subtree;
            nodes_in_last_level
                = nodes_in_last_level - last_level_nodes;
        }
        return Root;
    }
}
 
// This Code is contributed by Tapesh(tapeshdua420)

Output

1  2  5  6  7  3  8  9  4  

Time Complexity: O(M)
Auxiliary Space: O(M) for building the tree


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