Find sum of all nodes of the given perfect binary tree
Given a positive integer L which represents the number of levels in a perfect binary tree. Given that the leaf nodes in this perfect binary tree are numbered starting from 1 to n, where n is the number of leaf nodes. And the parent node is the sum of the two child nodes. Our task is to write a program to print the sum of all of the nodes of this perfect binary tree.
Examples:
Input : L = 3 Output : 30 Explanation : Tree will be - 10 / \ 3 7 / \ / \ 1 2 3 4 Input : L = 2 Output : 6 Explanation : Tree will be - 3 / \ 1 2
Naive Approach: The simplest solution is to first generate the value of all of the nodes of the perfect binary tree and then calculate the sum of all of the nodes. We can first generate all of the leaf nodes and then proceed in the bottom-up fashion to generate rest of the nodes. We know that in a perfect binary tree, the number of leaf nodes can be given by 2L-1, where L is the number of levels. The number of nodes in a perfect binary tree as we move upward from the bottom will get decreased by half.
Below is the implementation of above idea:
C++
#include <bits/stdc++.h> using namespace std; // function to find sum of all of the nodes // of given perfect binary tree int sumNodes( int l) { // no of leaf nodes int leafNodeCount = pow (2, l - 1); // list of vector to store nodes of // all of the levels vector< int > vec[l]; // store the nodes of last level // i.e., the leaf nodes for ( int i = 1; i <= leafNodeCount; i++) vec[l - 1].push_back(i); // store nodes of rest of the level // by moving in bottom-up manner for ( int i = l - 2; i >= 0; i--) { int k = 0; // loop to calculate values of parent nodes // from the children nodes of lower level while (k < vec[i + 1].size() - 1) { // store the value of parent node as // sum of children nodes vec[i].push_back(vec[i + 1][k] + vec[i + 1][k + 1]); k += 2; } } int sum = 0; // traverse the list of vector // and calculate the sum for ( int i = 0; i < l; i++) { for ( int j = 0; j < vec[i].size(); j++) sum += vec[i][j]; } return sum; } // Driver Code int main() { int l = 3; cout << sumNodes(l); return 0; } |
Java
// Java program to implement // the above approach import java.util.*; class GFG { // function to find sum of // all of the nodes of given // perfect binary tree static int sumNodes( int l) { // no of leaf nodes int leafNodeCount = ( int )Math.pow( 2 , l - 1 ); // list of vector to store // nodes of all of the levels Vector<Vector <Integer>> vec = new Vector<Vector <Integer>>(); //initialize for ( int i = 1 ; i <= l; i++) vec.add( new Vector<Integer>()); // store the nodes of last level // i.e., the leaf nodes for ( int i = 1 ; i <= leafNodeCount; i++) vec.get(l - 1 ).add(i); // store nodes of rest of // the level by moving in // bottom-up manner for ( int i = l - 2 ; i >= 0 ; i--) { int k = 0 ; // loop to calculate values // of parent nodes from the // children nodes of lower level while (k < vec.get(i + 1 ).size() - 1 ) { // store the value of parent // node as sum of children nodes vec.get(i).add(vec.get(i + 1 ).get(k) + vec.get(i + 1 ).get(k + 1 )); k += 2 ; } } int sum = 0 ; // traverse the list of vector // and calculate the sum for ( int i = 0 ; i < l; i++) { for ( int j = 0 ; j < vec.get(i).size(); j++) sum += vec.get(i).get(j); } return sum; } // Driver Code public static void main(String args[]) { int l = 3 ; System.out.println(sumNodes(l)); } } // This code is contributed // by Arnab Kundu |
Python3
# Python3 program to implement the # above approach # function to find Sum of all of the # nodes of given perfect binary tree def SumNodes(l): # no of leaf nodes leafNodeCount = pow ( 2 , l - 1 ) # list of vector to store nodes of # all of the levels vec = [[] for i in range (l)] # store the nodes of last level # i.e., the leaf nodes for i in range ( 1 , leafNodeCount + 1 ): vec[l - 1 ].append(i) # store nodes of rest of the level # by moving in bottom-up manner for i in range (l - 2 , - 1 , - 1 ): k = 0 # loop to calculate values of parent nodes # from the children nodes of lower level while (k < len (vec[i + 1 ]) - 1 ): # store the value of parent node as # Sum of children nodes vec[i].append(vec[i + 1 ][k] + vec[i + 1 ][k + 1 ]) k + = 2 Sum = 0 # traverse the list of vector # and calculate the Sum for i in range (l): for j in range ( len (vec[i])): Sum + = vec[i][j] return Sum # Driver Code if __name__ = = '__main__' : l = 3 print (SumNodes(l)) # This code is contributed by PranchalK |
C#
using System; using System.Collections.Generic; // C# program to implement // the above approach public class GFG { // function to find sum of // all of the nodes of given // perfect binary tree public static int sumNodes( int l) { // no of leaf nodes int leafNodeCount = ( int )Math.Pow(2, l - 1); // list of vector to store // nodes of all of the levels List<List< int >> vec = new List<List< int >>(); //initialize for ( int i = 1; i <= l; i++) { vec.Add( new List< int >()); } // store the nodes of last level // i.e., the leaf nodes for ( int i = 1; i <= leafNodeCount; i++) { vec[l - 1].Add(i); } // store nodes of rest of // the level by moving in // bottom-up manner for ( int i = l - 2; i >= 0; i--) { int k = 0; // loop to calculate values // of parent nodes from the // children nodes of lower level while (k < vec[i + 1].Count - 1) { // store the value of parent // node as sum of children nodes vec[i].Add(vec[i + 1][k] + vec[i + 1][k + 1]); k += 2; } } int sum = 0; // traverse the list of vector // and calculate the sum for ( int i = 0; i < l; i++) { for ( int j = 0; j < vec[i].Count; j++) { sum += vec[i][j]; } } return sum; } // Driver Code public static void Main( string [] args) { int l = 3; Console.WriteLine(sumNodes(l)); } } // This code is contributed by Shrikant13 |
Javascript
<script> // Javascript program to implement the above approach // function to find sum of // all of the nodes of given // perfect binary tree function sumNodes(l) { // no of leaf nodes let leafNodeCount = Math.pow(2, l - 1); // list of vector to store // nodes of all of the levels let vec = []; //initialize for (let i = 1; i <= l; i++) { vec.push([]); } // store the nodes of last level // i.e., the leaf nodes for (let i = 1; i <= leafNodeCount; i++) { vec[l - 1].push(i); } // store nodes of rest of // the level by moving in // bottom-up manner for (let i = l - 2; i >= 0; i--) { let k = 0; // loop to calculate values // of parent nodes from the // children nodes of lower level while (k < vec[i + 1].length - 1) { // store the value of parent // node as sum of children nodes vec[i].push(vec[i + 1][k] + vec[i + 1][k + 1]); k += 2; } } let sum = 0; // traverse the list of vector // and calculate the sum for (let i = 0; i < l; i++) { for (let j = 0; j < vec[i].length; j++) { sum += vec[i][j]; } } return sum; } let l = 3; document.write(sumNodes(l)); // This code is contributed by mukesh07. </script> |
30
Time Complexity: O(n), where n is the total number of nodes in the perfect binary tree.
Auxiliary Space: O(l)
Efficient Approach: An efficient approach is to observe that we only need to find the sum of all of the nodes. We can easily get the sum of all nodes at the last level using the formula of sum of first n natural numbers. Also, it can be seen that, as it is a perfect binary tree and parent nodes will be the sum of children nodes so the sum of nodes at all of the levels will be same. Therefore, we just need to find the sum of nodes at last level and multiply it by the total number of levels.
Below is the implementation of above idea:
C++
#include <bits/stdc++.h> using namespace std; // function to find sum of all of the nodes // of given perfect binary tree int sumNodes( int l) { // no of leaf nodes int leafNodeCount = pow (2, l - 1); int sumLastLevel = 0; // sum of nodes at last level sumLastLevel = (leafNodeCount * (leafNodeCount + 1)) / 2; // sum of all nodes int sum = sumLastLevel * l; return sum; } // Driver Code int main() { int l = 3; cout << sumNodes(l); return 0; } |
Java
// Java code to find sum of all nodes // of the given perfect binary tree import java.io.*; import java.lang.Math; class GFG { // function to find sum of // all of the nodes of given // perfect binary tree static double sumNodes( int l) { // no of leaf nodes double leafNodeCount = Math.pow( 2 , l - 1 ); double sumLastLevel = 0 ; // sum of nodes at last level sumLastLevel = (leafNodeCount * (leafNodeCount + 1 )) / 2 ; // sum of all nodes double sum = sumLastLevel * l; return sum; } // Driver Code public static void main (String[] args) { int l = 3 ; System.out.println(sumNodes(l)); } } // This code is contributed by // Anuj_{AJ_67} |
Python3
# function to find sum of all of the nodes # of given perfect binary tree import math def sumNodes(l): # no of leaf nodes leafNodeCount = math. pow ( 2 , l - 1 ); sumLastLevel = 0 ; # sum of nodes at last level sumLastLevel = ((leafNodeCount * (leafNodeCount + 1 )) / 2 ); # sum of all nodes sum = sumLastLevel * l; return int ( sum ); # Driver Code l = 3 ; print (sumNodes(l)); # This code is contributed by manishshaw |
C#
// C# code to find sum of all nodes // of the given perfect binary tree using System; using System.Collections.Generic; class GFG { // function to find sum of // all of the nodes of given // perfect binary tree static double sumNodes( int l) { // no of leaf nodes double leafNodeCount = Math.Pow(2, l - 1); double sumLastLevel = 0; // sum of nodes at last level sumLastLevel = (leafNodeCount * (leafNodeCount + 1)) / 2; // sum of all nodes double sum = sumLastLevel * l; return sum; } // Driver Code public static void Main() { int l = 3; Console.Write(sumNodes(l)); } } // This code is contributed by // Manish Shaw (manishshaw1) |
PHP
<?php // PHP code to find sum of all nodes // of the given perfect binary tree // function to find sum of // all of the nodes of given // perfect binary tree function sumNodes( $l ) { // no of leaf nodes $leafNodeCount = ( $l - 1) * ( $l - 1); $sumLastLevel = 0; // sum of nodes at last level $sumLastLevel = ( $leafNodeCount * ( $leafNodeCount + 1)) / 2; // sum of all nodes $sum = $sumLastLevel * $l ; return $sum ; } // Driver Code $l = 3; echo (sumNodes( $l )); // This code is contributed by // Manish Shaw (manishshaw1) ?> |
Javascript
<script> // Javascript code to find sum of all nodes // of the given perfect binary tree // function to find sum of // all of the nodes of given // perfect binary tree function sumNodes(l) { // no of leaf nodes let leafNodeCount = Math.pow(2, l - 1); let sumLastLevel = 0; // sum of nodes at last level sumLastLevel = (leafNodeCount * (leafNodeCount + 1)) / 2; // sum of all nodes let sum = sumLastLevel * l; return sum; } let l = 3; document.write(sumNodes(l)); // This code is contributed by divyeshrabadiya07. </script> |
30
Time Complexity: O(log(L)), due to pow()
Auxiliary Space: O(1)
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