Construct a Complete N-ary Tree from given Postorder Traversal

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

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

👁 Example 2

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 + . . . + N^H-1 = (N^H - 1)/(N - 1). 

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

The height H can be calculated as ceil[log_N( M*(N-1) +1)] - 1 because 
N-ary complete binary tree there can have at max (N^H+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 (N^H-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 N^H-1 nodes, then all levels in this subtree are full and the subtree has (N^H-1-1)/(N-1) + N^H-1 nodes.
   • Otherwise, the subtree will have (N^H-1-1)/(N-1) + L nodes where L is the number of nodes in the last level.
   • L can be calculated as L = M - (N^H - 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

2nd step: The remaining array will be broken into subtrees. For the first subtree H = 2.
5 > 3^2-1 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

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

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

👁 final structure of tree

After this step the tree is completely built.

Below is the implementation of the above approach:

1 2 5 6 7 3 8 9 4 

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