Find multiplication of sums of data of leaves at same levelss in C++


With respect of a given Binary Tree, return following value for it.

  • With respect of every level, calculate sum of all leaves if there are leaves at this level. Else ignore it.

  • Calculate multiplication of all sums and return it.


Root of following tree
     / \
    8   6



First level doesn’t have leaves. Second levelhas one leaf 8 and third level also has one leaf 10. So result is 8*10 = 80


Root of following tree
           /  \
           8   6
          / \   \
         9 7   10
           / \  / \
           2 12 5 11



First two levels don't have leaves. Thirdlevel has single leaf 9. Last level has fourleaves 2, 12, 5 and 11. So result is 9 * (2 + 12 + 5 + 11) = 270


With respect of one Simple Solution, we recursively calculate leaf sum for all level startingfrom top to bottom. After that multiply sums of levels which have leaves. Here, time complexity of this solutionwould be O(n^2).

Again With respect of an Efficient Solution, we implement Queue based level order traversal. Here, while doing the traversal, we separately process all different levels.With respect of every processed level, verify if it has a leaves. Inthis case, if it has then calculate sum of leaf nodes. At last, return product of all sums.


 Live Demo

/* Iterative C++ program to find sum of data of all leaves
of a binary tree on same level and then multiply sums
obtained of all levels. */
#include <bits/stdc++.h>
using namespace std;
// Shows a Binary Tree Node
struct Node1 {
   int data1;
   struct Node1 *left1, *right1;
// Shows helper function to check if a Node is leaf of tree
bool isLeaf(Node1* root1){
   return (!root1->left1 && !root1->right1);
/* Compute sum of all leaf Nodes at each level and returns
multiplication of sums */
int sumAndMultiplyLevelData(Node1* root1){
   // Here tree is empty
   if (!root1)
      return 0;
   int mul1 = 1; /* Used To store result */
   // Build an empty queue for level order tarversal
   queue<Node1*> q1;
   // Used to Enqueue Root and initialize height
   // Perform level order traversal of tree
   while (1) {
      // NodeCount1 (queue size) indicates number of Nodes
      // at current lelvel.
      int NodeCount1 = q1.size();
      // Now if there are no Nodes at current level, we are done
      if (NodeCount1 == 0)
      // Used to initialize leaf sum for current level
         int levelSum1 = 0;
      // Shows a boolean variable to indicate if found a leaf
      // Node at current level or not
      bool leafFound1 = false;
      // Used to Dequeue all Nodes of current level and Enqueue
      // Nodes of next level
      while (NodeCount1 > 0) {
         // Process next Node of current level
         Node1* Node1 = q1.front();
         /* Now if Node is a leaf, update sum at the level */
         if (isLeaf(Node1)) {
            leafFound1 = true;
            levelSum1 += Node1->data1;
         // Add children of Node
         if (Node1->left1 != NULL)
         if (Node1->right1 != NULL)
      // Now if we found at least one leaf, we multiply
      // result with level sum.
      if (leafFound1)
         mul1 *= levelSum1;
   return mul1; // Here, return result
//Shows utility function to create a new tree Node
Node1* newNode(int data1){
   Node1* temp1 = new Node1;
   temp1->data1 = data1;
   temp1->left1 = temp1->right1 = NULL;
   return temp1;
// Driver program to test above functions
int main(){
   Node1* root1 = newNode(3);
   root1->left1 = newNode(8);
   root1->right1 = newNode(6);
   root1->left1->right1 = newNode(7);
   root1->left1->left1 = newNode(9);
   root1->left1->right1->left1 = newNode(2);
   root1->left1->right1->right1 = newNode(12);
   root1->right1->right1 = newNode(10);
   root1->right1->right1->left1 = newNode(5);
   root1->right1->right1->right1 = newNode(11);
   cout << "Final product value = "
   << sumAndMultiplyLevelData(root1) <<endl;
   return 0;


Final product value = 270

Updated on: 25-Jul-2020


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