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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 using namespace std; // function to find sum of all of the nodes// of given perfect binary treeint 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 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 Codeint main(){    int l = 3;     cout << sumNodes(l);     return 0;}

Java

 // Java program to implement// the above approachimport java.util.*;class GFG{ // function to find sum of// all of the nodes of given// perfect binary treestatic 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> vec = new Vector>();         //initialize    for (int i = 1; i <= l; i++)    vec.add(new Vector());         // 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 Codepublic 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 treedef 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 Codeif __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 approachpublic 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> vec = new List>();     //initialize    for (int i = 1; i <= l; i++)    {    vec.Add(new List());    }     // 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



Output

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 using namespace std; // function to find sum of all of the nodes// of given perfect binary treeint 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 Codeint main(){    int l = 3;    cout << sumNodes(l);    return 0;}

Java

 // Java code to find sum of all nodes// of the given perfect binary treeimport 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 treeimport 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 Codel = 3;print (sumNodes(l)); # This code is contributed by manishshaw

C#

 // C# code to find sum of all nodes// of the given perfect binary treeusing 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)



Javascript



Output

30

Time Complexity: O(log(L)), due to pow()
Auxiliary Space: O(1)