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Smallest value in each level of Binary Tree

Given a binary tree containing n nodes, the task is to print the minimum elements in each level of the binary tree.

Examples: 

Input : 
            7
         /    \
        6       5
       / \     / \
      4  3     2  1  
       
Output : 
Every level minimum is
level 0 min is = 7
level 1 min is = 5
level 2 min is = 1

Input :  
            7
          /    \
        16       1
       / \      
      4   13    

Output :
Every level minimum is
level 0 min is = 7
level 1 min is = 1
level 2 min is = 4

Method 1: Using In-order traversal 

Approach:- The idea is to recursively traverse trees in an in-order fashion. The root is considered to be at the zeroth level. First, find the height of the tree and store it into res. res array store every smallest element in each level of a binary tree.

Below is the implementation to find the smallest value on each level of the Binary Tree. 




// CPP program to print smallest element
// in each level of binary tree.
#include <bits/stdc++.h>
using namespace std;
 
// A Binary Tree Node
struct Node {
    int data;
    struct Node *left, *right;
};
 
// return height of tree
int heightoftree(Node* root)
{
 
    if (root == NULL)
        return 0;
 
    int left = heightoftree(root->left);
    int right = heightoftree(root->right);
 
    return ((left > right ? left : right) + 1);
}
 
// Inorder Traversal
// Search minimum element in each level and
// store it into vector array.
void printPerLevelMinimum(Node* root,
                  vector<int>& res, int level)
{
     
    if (root != NULL) {
 
        printPerLevelMinimum(root->left,
                              res, level + 1);
 
        if (root->data < res[level])
            res[level] = root->data;
 
        printPerLevelMinimum(root->right,
                              res, level + 1);
    }
}
 
void perLevelMinimumUtility(Node* root)
{
     
    // height of tree for the size of
    // vector array
    int n = heightoftree(root), i;
 
    // vector for store all minimum of
    // every level
    vector<int> res(n, __INT_MAX__);
 
    // save every level minimum using
    // inorder traversal
    printPerLevelMinimum(root, res, 0);
 
    // print every level minimum
    cout << "Every level minimum is\n";
    for (i = 0; i < n; i++) {
        cout << "level " << i <<" min is = "
                            << res[i] << "\n";
    }
}
 
// Utility function to create a new tree node
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
 
    return temp;
}
 
// Driver program to test above functions
int main()
{
 
    // Let us create binary tree shown
    // in above diagram
    Node* root = newNode(7);
    root->left = newNode(6);
    root->right = newNode(5);
    root->left->left = newNode(4);
    root->left->right = newNode(3);
    root->right->left = newNode(2);
    root->right->right = newNode(1);
 
    /*       7
         /  \
        6     5
       / \     / \
      4   3 2   1         */
    perLevelMinimumUtility(root);
 
    return 0;
}




// Java program to print smallest element
// in each level of binary tree.
import java.util.Arrays;
class GFG
{
static int INT_MAX = (int) 10e6;
 
// A Binary Tree Node
static class Node
{
    int data;
    Node left, right;
};
 
// return height of tree
static int heightoftree(Node root)
{
    if (root == null)
        return 0;
 
    int left = heightoftree(root.left);
    int right = heightoftree(root.right);
 
    return ((left > right ? left : right) + 1);
}
 
// Inorder Traversal
// Search minimum element in each level and
// store it into vector array.
static void printPerLevelMinimum(Node root,
                      int []res, int level)
{
    if (root != null)
    {
        printPerLevelMinimum(root.left,
                             res, level + 1);
 
        if (root.data < res[level])
            res[level] = root.data;
 
        printPerLevelMinimum(root.right,
                             res, level + 1);
    }
}
 
static void perLevelMinimumUtility(Node root)
{
     
    // height of tree for the size of
    // vector array
    int n = heightoftree(root), i;
 
    // vector for store all minimum of
    // every level
    int []res = new int[n];
    Arrays.fill(res, INT_MAX);
 
    // save every level minimum using
    // inorder traversal
    printPerLevelMinimum(root, res, 0);
 
    // print every level minimum
    System.out.print("Every level minimum is\n");
    for (i = 0; i < n; i++)
    {
        System.out.print("level " + i +
                         " min is = " +
                        res[i] + "\n");
    }
}
 
// Utility function to create a new tree node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
 
    return temp;
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Let us create binary tree shown
    // in above diagram
    Node root = newNode(7);
    root.left = newNode(6);
    root.right = newNode(5);
    root.left.left = newNode(4);
    root.left.right = newNode(3);
    root.right.left = newNode(2);
    root.right.right = newNode(1);
 
    /*     7
        / \
        6     5
    / \     / \
    4 3 2 1         */
    perLevelMinimumUtility(root);
}
}
 
// This code is contributed by PrinciRaj1992




# Python3 program to print
# smallest element in each
# level of binary tree.
INT_MAX = 1000006
  
# A Binary Tree Node
class Node:
   
    def __init__(self,
                 data):
       
        self.data = data
        self.left = None
        self.right = None   
  
# return height of tree
def heightoftree(root):
  
    if (root == None):
        return 0;
  
    left = heightoftree(root.left);
    right = heightoftree(root.right);
     
    return ((left if left > right
                  else right) + 1);
 
  
# Inorder Traversal
# Search minimum element in
# each level and store it
# into vector array.
def printPerLevelMinimum(root,
                         res, level):
      
    if (root != None):
        res = printPerLevelMinimum(root.left,
                                   res, level + 1);
        if (root.data < res[level]):
            res[level] = root.data;
             
        res = printPerLevelMinimum(root.right,
                                   res, level + 1);
    return res
  
def perLevelMinimumUtility(root):
      
    # height of tree for the
    # size of vector array
    n = heightoftree(root)
    i = 0
  
    # vector for store all
    # minimum of every level
    res = [INT_MAX for i in range(n)]
  
    # save every level minimum
    # using inorder traversal
    res = printPerLevelMinimum(root,
                               res, 0);
  
    # print every level minimum
    print("Every level minimum is")
     
    for i in range(n):
        print('level ' + str(i) +
              ' min is = ' + str(res[i]))       
  
# Utility function to create
# a new tree node
def newNode(data):
 
    temp = Node(data)
    return temp;
 
# Driver code
if __name__ == "__main__":
     
    # Let us create binary
    # tree shown in below
    # diagram
    root = newNode(7);
    root.left = newNode(6);
    root.right = newNode(5);
    root.left.left = newNode(4);
    root.left.right = newNode(3);
    root.right.left = newNode(2);
    root.right.right = newNode(1);
  
    '''    7
         /  \
        6     5
       / \     / \
      4   3 2   1         '''
       
    perLevelMinimumUtility(root);
  
# This code is contributed by Rutvik_56




// C# program to print smallest element
// in each level of binary tree.
using System;
 
class GFG
{
static int INT_MAX = (int) 10e6;
 
// A Binary Tree Node
public class Node
{
    public int data;
    public Node left, right;
};
 
// return height of tree
static int heightoftree(Node root)
{
    if (root == null)
        return 0;
 
    int left = heightoftree(root.left);
    int right = heightoftree(root.right);
 
    return ((left > right ? left : right) + 1);
}
 
// Inorder Traversal
// Search minimum element in each level and
// store it into vector array.
static void printPerLevelMinimum(Node root,
                                 int []res,
                                 int level)
{
    if (root != null)
    {
        printPerLevelMinimum(root.left,
                             res, level + 1);
 
        if (root.data < res[level])
            res[level] = root.data;
 
        printPerLevelMinimum(root.right,
                             res, level + 1);
    }
}
 
static void perLevelMinimumUtility(Node root)
{
     
    // height of tree for the size of
    // vector array
    int n = heightoftree(root), i;
 
    // vector for store all minimum of
    // every level
    int []res = new int[n];
    for (i = 0; i < n; i++)
        res[i] = INT_MAX;
 
    // save every level minimum using
    // inorder traversal
    printPerLevelMinimum(root, res, 0);
 
    // print every level minimum
    Console.Write("Every level minimum is\n");
    for (i = 0; i < n; i++)
    {
        Console.Write("level " + i +
                      " min is = " +
                     res[i] + "\n");
    }
}
 
// Utility function to create a new tree node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
 
    return temp;
}
 
// Driver Code
public static void Main(String[] args)
{
 
    // Let us create binary tree shown
    // in above diagram
    Node root = newNode(7);
    root.left = newNode(6);
    root.right = newNode(5);
    root.left.left = newNode(4);
    root.left.right = newNode(3);
    root.right.left = newNode(2);
    root.right.right = newNode(1);
 
    /*     7
        / \
        6     5
    / \     / \
    4 3 2 1         */
    perLevelMinimumUtility(root);
}
}
 
// This code is contributed by Princi Singh




<script>
 
    // JavaScript program to print smallest element
    // in each level of binary tree.
     
    let INT_MAX = 10e6;
  
    // A Binary Tree Node
    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
 
    // return height of tree
    function heightoftree(root)
    {
        if (root == null)
            return 0;
 
        let left = heightoftree(root.left);
        let right = heightoftree(root.right);
 
        return ((left > right ? left : right) + 1);
    }
 
    // Inorder Traversal
    // Search minimum element in each level and
    // store it into vector array.
    function printPerLevelMinimum(root, res, level)
    {
        if (root != null)
        {
            printPerLevelMinimum(root.left, res, level + 1);
 
            if (root.data < res[level])
                res[level] = root.data;
 
            printPerLevelMinimum(root.right, res, level + 1);
        }
    }
 
    function perLevelMinimumUtility(root)
    {
 
        // height of tree for the size of
        // vector array
        let n = heightoftree(root), i;
 
        // vector for store all minimum of
        // every level
        let res = new Array(n);
        res.fill(INT_MAX);
 
        // save every level minimum using
        // inorder traversal
        printPerLevelMinimum(root, res, 0);
 
        // print every level minimum
        document.write("Every level minimum is" + "</br>");
        for (i = 0; i < n; i++)
        {
            document.write("level " + i +
                             " min is = " +
                            res[i] + "</br>");
        }
    }
 
    // Utility function to create a new tree node
    function newNode(data)
    {
        let temp = new Node(data);
        temp.data = data;
        temp.left = temp.right = null;
 
        return temp;
    }
     
    // Let us create binary tree shown
    // in above diagram
    let root = newNode(7);
    root.left = newNode(6);
    root.right = newNode(5);
    root.left.left = newNode(4);
    root.left.right = newNode(3);
    root.right.left = newNode(2);
    root.right.right = newNode(1);
  
    /*     7
        / \
        6     5
    / \     / \
    4 3 2 1         */
    perLevelMinimumUtility(root);
 
</script>

Output
Every level minimum is
level 0 min is = 7
level 1 min is = 5
level 2 min is = 1

Time Complexity: O(N) where N is the number of nodes
Auxiliary Space: O(h), where h is the height of given binary tree.

Method 2: Using level order Traversal 

Approach:- The idea is to perform iterative level order traversal of the binary tree using a queue. While traversing keep min variable which stores the minimum element of the current level of the tree being processed. When the level is completely traversed, print that min value. 




// CPP program to print minimum element
// in each level of binary tree.
#include <iostream>
#include <queue>
#include <vector>
#define INT_MAX 10e6
using namespace std;
 
// A Binary Tree Node
struct Node {
    int data;
    struct Node *left, *right;
};
 
// return height of tree
int heightoftree(Node* root)
{
 
    if (root == NULL)
        return 0;
 
    int left = heightoftree(root->left);
    int right = heightoftree(root->right);
 
    return ((left > right ? left : right) + 1);
}
 
// Iterative method to find every level
// minimum element of Binary Tree
void printPerLevelMinimum(Node* root)
{
 
    // Base Case
    if (root == NULL)
        return ;
 
    // Create an empty queue for
    // level order traversal
    queue<Node*> q;
 
    // push the root for Change the level
    q.push(root);
 
    // for go level by level
    q.push(NULL);
 
    int min = INT_MAX;
    // for check the level
    int level = 0;
 
    while (q.empty() == false) {
 
        // Get top of queue
        Node* node = q.front();
        q.pop();
 
        // if node == NULL (Means this is
        // boundary between two levels)
        if (node == NULL) {
 
            cout << "level " << level <<
             " min is = " << min << "\n";
 
            // here queue is empty represent
            // no element in the actual
            // queue
            if (q.empty())
                break;
 
            q.push(NULL);
 
            // increment level
            level++;
 
            // Reset min for next level
            // minimum value
            min = INT_MAX;
 
            continue;
        }
 
        // get Minimum in every level
        if (min > node->data)
            min = node->data;
 
        /* Enqueue left child */
        if (node->left != NULL) {
            q.push(node->left);
        }
 
        /*Enqueue right child */
        if (node->right != NULL) {
            q.push(node->right);
        }
    }
}
 
// Utility function to create a
// new tree node
Node* newNode(int data)
{
     
    Node* temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
     
    return temp;
}
 
// Driver program to test above functions
int main()
{
     
    // Let us create binary tree shown
    // in above diagram
    Node* root = newNode(7);
    root->left = newNode(6);
    root->right = newNode(5);
    root->left->left = newNode(4);
    root->left->right = newNode(3);
    root->right->left = newNode(2);
    root->right->right = newNode(1);
 
    /*      7
        /  \
       6    5
      / \  / \
     4  3 2   1         */
 
    cout << "Every Level minimum is"
        << "\n";
         
    printPerLevelMinimum(root);
     
    return 0;
}




// JAVA program to print minimum element
// in each level of binary tree.
import java.util.*;
 
class GFG
{
 
// A Binary Tree Node
static class Node
{
    int data;
    Node left, right;
};
 
// return height of tree
static int heightoftree(Node root)
{
 
    if (root == null)
        return 0;
 
    int left = heightoftree(root.left);
    int right = heightoftree(root.right);
 
    return ((left > right ? left : right) + 1);
}
 
// Iterative method to find every level
// minimum element of Binary Tree
static void printPerLevelMinimum(Node root)
{
 
    // Base Case
    if (root == null)
        return ;
 
    // Create an empty queue for
    // level order traversal
    Queue<Node> q = new LinkedList<Node>();
 
    // push the root for Change the level
    q.add(root);
 
    // for go level by level
    q.add(null);
 
    int min = Integer.MAX_VALUE;
    // for check the level
    int level = 0;
 
    while (q.isEmpty() == false)
    {
 
        // Get top of queue
        Node node = q.peek();
        q.remove();
 
        // if node == null (Means this is
        // boundary between two levels)
        if (node == null)
        {
 
            System.out.print("level " + level +
            " min is = " + min+ "\n");
 
            // here queue is empty represent
            // no element in the actual
            // queue
            if (q.isEmpty())
                break;
 
            q.add(null);
 
            // increment level
            level++;
 
            // Reset min for next level
            // minimum value
            min = Integer.MAX_VALUE;
 
            continue;
        }
 
        // get Minimum in every level
        if (min > node.data)
            min = node.data;
 
        /* Enqueue left child */
        if (node.left != null)
        {
            q.add(node.left);
        }
 
        /*Enqueue right child */
        if (node.right != null)
        {
            q.add(node.right);
        }
    }
}
 
// Utility function to create a
// new tree node
static Node newNode(int data)
{
     
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
     
    return temp;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Let us create binary tree shown
    // in above diagram
    Node root = newNode(7);
    root.left = newNode(6);
    root.right = newNode(5);
    root.left.left = newNode(4);
    root.left.right = newNode(3);
    root.right.left = newNode(2);
    root.right.right = newNode(1);
 
    /*     7
        / \
    6 5
    / \ / \
    4 3 2 1         */
 
    System.out.print("Every Level minimum is"
    + "\n");
         
    printPerLevelMinimum(root);
     
}
}
 
// This code is contributed by Rajput-Ji




# Python3 program to print minimum element
# in each level of binary tree.
 
# Importing Queue
from queue import Queue
 
# Utility class to create a
# new tree node
class newNode:
    def __init__(self, data):
        self.data = data
        self.left = self.right = None
     
# return height of tree p
def heightoftree(root):
 
    if (root == None):
        return 0
 
    left = heightoftree(root.left)
    right = heightoftree(root.right)
    if left > right:
        return left + 1
    else:
        return right + 1
         
# Iterative method to find every level
# minimum element of Binary Tree
def printPerLevelMinimum(root):
 
    # Base Case
    if (root == None):
        return
 
    # Create an empty queue for
    # level order traversal
    q = Queue()
 
    # put the root for Change the level
    q.put(root)
 
    # for go level by level
    q.put(None)
 
    Min = 9999999999999
     
    # for check the level
    level = 0
 
    while (q.empty() == False):
 
        # Get top of queue
        node = q.queue[0]
        q.get()
 
        # if node == None (Means this is
        # boundary between two levels)
        if (node == None):
 
            print("level", level, "min is =", Min)
 
            # here queue is empty represent
            # no element in the actual
            # queue
            if (q.empty()):
                break
 
            q.put(None)
 
            # increment level
            level += 1
 
            # Reset min for next level
            # minimum value
            Min = 999999999999
 
            continue
 
        # get Minimum in every level
        if (Min > node.data):
            Min = node.data
 
        # Enqueue left child
        if (node.left != None):
            q.put(node.left)
 
        #Enqueue right child
        if (node.right != None):
            q.put(node.right)
 
# Driver Code
if __name__ == '__main__':
     
    # Let us create binary tree shown
    # in above diagram
    root = newNode(7)
    root.left = newNode(6)
    root.right = newNode(5)
    root.left.left = newNode(4)
    root.left.right = newNode(3)
    root.right.left = newNode(2)
    root.right.right = newNode(1)
 
    #     7
    # / \
    # 6 5
    # / \ / \
    # 4 3 2 1        
    print("Every Level minimum is")
         
    printPerLevelMinimum(root)
 
# This code is contributed by PranchalK




// C# program to print minimum element
// in each level of binary tree.
using System;
using System.Collections.Generic;
 
class GFG
{
 
// A Binary Tree Node
class Node
{
    public int data;
    public Node left, right;
};
 
// return height of tree
static int heightoftree(Node root)
{
 
    if (root == null)
        return 0;
 
    int left = heightoftree(root.left);
    int right = heightoftree(root.right);
 
    return ((left > right ? left : right) + 1);
}
 
// Iterative method to find every level
// minimum element of Binary Tree
static void printPerLevelMinimum(Node root)
{
 
    // Base Case
    if (root == null)
        return;
 
    // Create an empty queue for
    // level order traversal
    Queue<Node> q = new Queue<Node>();
 
    // push the root for Change the level
    q.Enqueue(root);
 
    // for go level by level
    q.Enqueue(null);
 
    int min = int.MaxValue;
    // for check the level
    int level = 0;
 
    while (q.Count != 0)
    {
 
        // Get top of queue
        Node node = q.Peek();
        q.Dequeue();
 
        // if node == null (Means this is
        // boundary between two levels)
        if (node == null)
        {
 
            Console.Write("level " + level +
                          " min is = " + min + "\n");
 
            // here queue is empty represent
            // no element in the actual
            // queue
            if (q.Count == 0)
                break;
 
            q.Enqueue(null);
 
            // increment level
            level++;
 
            // Reset min for next level
            // minimum value
            min = int.MaxValue;
 
            continue;
        }
 
        // get Minimum in every level
        if (min > node.data)
            min = node.data;
 
        /* Enqueue left child */
        if (node.left != null)
        {
            q.Enqueue(node.left);
        }
 
        /*Enqueue right child */
        if (node.right != null)
        {
            q.Enqueue(node.right);
        }
    }
}
 
// Utility function to create a
// new tree node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
     
    return temp;
}
 
// Driver code
public static void Main(String[] args)
{
     
    // Let us create binary tree shown
    // in above diagram
    Node root = newNode(7);
    root.left = newNode(6);
    root.right = newNode(5);
    root.left.left = newNode(4);
    root.left.right = newNode(3);
    root.right.left = newNode(2);
    root.right.right = newNode(1);
 
    /* 7
        / \
    6 5
    / \ / \
    4 3 2 1     */
 
    Console.Write("Every Level minimum is" + "\n");
         
    printPerLevelMinimum(root);
}
}
 
// This code is contributed by PrinciRaj1992




<script>
    // Javascript program to print minimum element
    // in each level of binary tree.
     
    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
 
    // return height of tree
    function heightoftree(root)
    {
 
        if (root == null)
            return 0;
 
        let left = heightoftree(root.left);
        let right = heightoftree(root.right);
 
        return ((left > right ? left : right) + 1);
    }
 
    // Iterative method to find every level
    // minimum element of Binary Tree
    function printPerLevelMinimum(root)
    {
 
        // Base Case
        if (root == null)
            return ;
 
        // Create an empty queue for
        // level order traversal
        let q = [];
 
        // push the root for Change the level
        q.push(root);
 
        // for go level by level
        q.push(null);
 
        let min = Number.MAX_VALUE;
        // for check the level
        let level = 0;
 
        while (q.length > 0)
        {
 
            // Get top of queue
            let node = q[0];
            q.shift();
 
            // if node == null (Means this is
            // boundary between two levels)
            if (node == null)
            {
 
                document.write("level " + level +
                " min is = " + min+ "</br>");
 
                // here queue is empty represent
                // no element in the actual
                // queue
                if (q.length == 0)
                    break;
 
                q.push(null);
 
                // increment level
                level++;
 
                // Reset min for next level
                // minimum value
                min = Number.MAX_VALUE;
 
                continue;
            }
 
            // get Minimum in every level
            if (min > node.data)
                min = node.data;
 
            /* Enqueue left child */
            if (node.left != null)
            {
                q.push(node.left);
            }
 
            /*Enqueue right child */
            if (node.right != null)
            {
                q.push(node.right);
            }
        }
    }
 
    // Utility function to create a
    // new tree node
    function newNode(data)
    {
 
        let temp = new Node(data);
        return temp;
    }
     
    // Let us create binary tree shown
    // in above diagram
    let root = newNode(7);
    root.left = newNode(6);
    root.right = newNode(5);
    root.left.left = newNode(4);
    root.left.right = newNode(3);
    root.right.left = newNode(2);
    root.right.right = newNode(1);
  
    /*     7
        / \
    6 5
    / \ / \
    4 3 2 1         */
  
    document.write("Every level minimum is"
    + "</br>");
          
    printPerLevelMinimum(root);
 
// This code is contributed by decode2207.
</script>

Output
Every Level minimum is
level 0 min is = 7
level 1 min is = 5
level 2 min is = 1

Time Complexity: O(N) where n is the number of nodes in the binary tree.
Auxiliary Space: O(N) where n is the number of nodes in the binary tree.


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