1.
Introduction
2.
Problem Statement
3.
Problem Example
4.
Brute force approach
5.
Pseudocode
6.
Implementation
6.1.
Output
7.
Complexity analysis
8.
Optimized approach
9.
Pseudocode
10.
Implementation
10.1.
Output:
11.
Complexity analysis
12.
12.1.
What is the difference between a binary tree and a binary search tree?
12.2.
How do we allocate space for a new node?
12.3.
How do we check for a leaf node?
13.
Conclusion
Last Updated: Mar 27, 2024

# Sum of Leaf Nodes at Minimum Level

Md Yawar
0 upvote
Master Power BI using Netflix Data
Speaker
Ashwin Goyal
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18 Jun, 2024 @ 01:30 PM

## Introduction

A tree is a heirarchial data structure that is described as a collection of nodes. The nodes are connected together to represent a hierarchy. It is a hierarchical structure since its pieces are placed at several levels. The tree is a popular topic asked in interviews, and one must practice questions to master this data structure. In this blog, we will talk about a tree-based problem in which we find the sum of the leaf nodes at minimum level.

## Problem Statement

You are given a binary tree with n nodes. The goal is to get the sum of leaf nodes at minimum level in the tree.

Note: Minimum level does not mean the lowest level of the tree. Here, minimum means the lesser level. For example, between level 3 and level 5, the minimum level is 3.

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## Problem Example

Output: 102

The tree's leaf nodes at the minimum level are 10, 19, 31, and 42. The sum of the leaf nodes at minimum level of this tree is 102.

Output: 6

The leaf nodes at the minimum level in the tree are 6, and the sum is 6.

## Brute force approach

Find the first level containing a leaf node by doing iterative level order traversal using a queue. After adding up all the leaf nodes at this level, stop traversing further.

## Pseudocode

• First, we create the binary tree.
• Then we will check if the binary tree is empty or not. If yes, we return 0. Else, carry on.
• Then we will check if the binary tree has only one element. If yes, we return the root of the tree. Else carry on.
• We will create a queue for the level order traversal of the tree.
• Then we set the flag to 0 and check for the leaf nodes.
• When we encounter the leaf nodes at the minimum level, we accumulate them to the sum.
• After getting the required sum, we set the flag value to 1 to indicate the minimum level has been reached, and no further check is needed.
• Return the sum.

## Implementation

``````// implementation in C++ to find the sum of leaf nodes at minimum level
#include <bits/stdc++.h>
using namespace std;

// structure of the node
struct Node {
int dat;
Node *left, *right;
};

// function to get a new node
Node* getNode(int dat)
{
//allocate space
Node* newNode = (Node*)malloc(sizeof(Node));

// inserting the data
newNode->dat = dat;
newNode->left = newNode->right = NULL;
return newNode;
}

// function to find the sum of leaf nodes at minimum level
int sumOfLeafNodesAtTheMinimumLevel(Node* root)
{
// if tree is empty
if (!root)
return 0;

// if there is only one node
if (!root->left && !root->right)
return root->dat;

// queue used for level order traversal
queue<Node*> queue;
int sum = 0;
bool flag = 0;

// push root node in the queue 'queue'
queue.push(root);

while (flag == 0) {

// count number of nodes in the
// current level
int nc = queue.size();

// traverse the current level nodes
while (nc--) {

// get front element from 'queue'
Node* top = queue.front();
queue.pop();

// if it is a leaf node
if (!top->left && !top->right) {

// accumulate dat to 'sum'
sum += top->dat;

// set flag to 1 to signify minimum level for leaf nodes has been encountered
flag = 1;
}

else {

// if top's left and right child exists, push them to 'queue'
if (top->left)
queue.push(top->left);
if (top->right)
queue.push(top->right);
}
}
}

// required sum
return sum;
}

// Driver code
int main()
{
// binary tree creation
Node* root = getNode(1);
root->left = getNode(3);
root->right = getNode(4);
root->left->left = getNode(6);
root->left->right = getNode(5);
root->right->left = getNode(8);
root->right->right = getNode(3);
root->left->right->left = getNode(11);
root->right->left->right = getNode(10);

cout << "Sum of leaf nodes at minimum level = "
<< sumOfLeafNodesAtTheMinimumLevel(root);

return 0;
}``````

#### Output

Sum of leaf nodes at minimum level = 9

## Complexity analysis

Time complexity: O(n), n is the number of elements in the tree. It is of the order of O(n) because we have to do an iterative tree traversal.

Space complexity: O(n), n is the number of elements present in the tree. It is of the order of O(n) because we have to allocate new space for the tree.

## Optimized approach

In this approach, we'll apply binary tree traversal. We will use a variable in to monitor the levels of each node. We will also use a hashmap, whose key is our level value. The value part of the hashmap will be used as a vector to store the node's data for that specific level. If the current node is a leaf node, we will push into the map with the key as level and the node's contents into a vector for each recursive iteration, increasing the level variable for each one. Once we have our map, all we need to do is add the vector of the first element.

## Pseudocode

• First, we create the binary tree.
• Then we create a map as map <int, vector <int>>. Here the key will hold the level value, and the data part will hold the leaf nodes at that level.
• We traverse the tree recursively and check for the leaf nodes at each level. If we encounter a leaf node, we push the node into the vector.
• After traversing the tree, we check for the minimum level of the tree and get the sum for the vector at that level.
• Return the sum.

## Implementation

``````// implementation in C++ to find the sum of leaf nodes at minimum level
#include <bits/stdc++.h>
using namespace std;

struct Node {
int dat;
Node *left, *right;
};

// function to get a new node
Node* getNode(int dat)
{
// allocate space
Node* newNode = (Node*)malloc(sizeof(Node));

// put in the data
newNode->dat = dat;
newNode->left = newNode->right = NULL;
return newNode;
}

map <int, vector <int>> map_;
void solve(Node* root, int level) {

if(root == NULL)
return;

if(root->left == NULL && root->right == NULL)
map_[level].push_back(root->dat);

solve(root->left, level+1);
solve(root->right, level+1);
}
int sumOfLeafNodesAtTheMinimumLevel(Node *root)
{
solve(root, 0);
int sum = 0;
for(auto i:map_) {
for(auto j:i.second) {
sum += j;
}
return sum;
}
}

int main() {
// binary tree creation
Node* root = getNode(1);
root->left = getNode(4);
root->right = getNode(3);
root->left->right = getNode(2);
root->left->left = getNode(6);
root->right->right = getNode(11);
root->right->left = getNode(8);
cout << "Sum of leaf nodes at minimum level = "
<< sumOfLeafNodesAtTheMinimumLevel(root);
return 0;
}``````

#### Output:

Sum of leaf nodes at minimum level = 27

## Complexity analysis

Time complexity: O(logn), where n is the number of elements in the tree. It is of the order of O(logn) because we have to recursively travel the tree.

Space complexity: O(n), n is the number of elements in the tree. It is of the order of O(n) because we have to allocate new space for the tree.

Check out this problem - Diameter Of Binary Tree

#### What is the difference between a binary tree and a binary search tree?

A binary tree is a type of tree in which the elements have at most 2 children. It does not follow any particular order. The elements in the binary search tree can also have at most 2 children, but it follows a particular order. The elements on the left side of a specific element are smaller than that element, and the elements on the right side of an element are bigger than that element.

#### How do we allocate space for a new node?

We can allocate space for a node using Node* newNode = (Node*)malloc(sizeof(Node)).

#### How do we check for a leaf node?

For a leaf node, we check for the children of the node. If both the left and right nodes of the parent node are NULL, it is a leaf node. We can check it by using if (!node->left && !node->right). If yes, then it is a leaf node.

## Conclusion

In this blog, we looked at a popular problem in which we find the sum of leaf nodes at minimum level.

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