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> |
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> |
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.