Find sum of all nodes of the given perfect binary tree
Last Updated :
26 Dec, 2022
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 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 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
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
<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>
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 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}
# 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# 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 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)
?>
<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>
Time Complexity: O(log(L)), due to pow()
Auxiliary Space: O(1)
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