Program to find tree level that has minimum sum in C++

Suppose we have a binary tree, the level of its root is 1, the level of its children is 2, and so on.We have to find the smallest level X such that the sum of all the values of nodes at level X is minimum. So if the tree is like −

Output will be 2 as the sum is 4 – 10 = -6, which is minimum.

To solve this, we will follow these steps −

• level := 1, sum := value of r, ansLevel := level, ansSum := sum

• define a queue q, insert given node r into q

• while q is not empty

  • capacity := size of q

  • increase level by 1, sum := 0

  • while capacity is not 0

    • node := front node from q, delete node from q

    • if right of node is valid, then sum := sum + value of right node, insert right

    • node into q

    • if left of node is valid, then sum := sum + value of left node, insert left node into q

    • decrease capacity by 1

  • if ansSum < sum, then ansSum := sum, ansLevel := level

• return ansLevel

Let us see the following implementation to get better understanding−

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class TreeNode{
   public:
      int val;
      TreeNode *left, *right;
      TreeNode(int data){
         val = data;
         left = NULL;
      right = NULL;
      }
};
class Solution {
   public:
   int solve(TreeNode* r) {
      int level = 1, sum = r->val;
      int ansLevel = level, ansSum = sum;
      queue <TreeNode*> q;
      q.push(r);
      while(!q.empty()){
         int capacity = q.size();
         level++;
         sum = 0;
         while(capacity--){
            TreeNode* node = q.front();
            q.pop();
            if(node->right){
               sum += node->right->val;
               q.push(node->right);
            }
            if(node->left){
               sum += node->left->val;
               q.push(node->left);
            }
         }
         if(ansSum>sum){
            ansSum = sum;
            ansLevel = level;
         }
      }
      return ansLevel;
   }
};
main(){
   TreeNode *root = new TreeNode(5);
   root->left = new TreeNode(4);
   root->right = new TreeNode(-10);
   root->left->right = new TreeNode(-2);
   root->right->left = new TreeNode(-7);
   root->right->right = new TreeNode(15);
   Solution ob;
   cout <<ob.solve(root);
}

Input

TreeNode *root = new TreeNode(5);
root->left = new TreeNode(4);
root->right = new TreeNode(-10);
root->left->right = new TreeNode(-2);
root->right->left = new TreeNode(-7);
root->right->right = new TreeNode(15);

Output

2