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.
C++
// CPP program to print smallest element// in each level of binary tree.#include <bits/stdc++.h>using namespace std;// A Binary Tree Nodestruct Node { int data; struct Node *left, *right;};// return height of treeint 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 nodeNode* newNode(int data){ Node* temp = new Node; temp->data = data; temp->left = temp->right = NULL; return temp;}// Driver program to test above functionsint 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
// 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 Nodestatic class Node { int data; Node left, right;};// return height of treestatic 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 nodestatic Node newNode(int data){ Node temp = new Node(); temp.data = data; temp.left = temp.right = null; return temp;}// Driver Codepublic 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
# Python3 program to print # smallest element in each # level of binary tree.INT_MAX = 1000006 # A Binary Tree Nodeclass Node: def __init__(self, data): self.data = data self.left = None self.right = None # return height of treedef 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 nodedef newNode(data): temp = Node(data) return temp;# Driver codeif __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#
// 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 Nodepublic class Node { public int data; public Node left, right;};// return height of treestatic 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 nodestatic Node newNode(int data){ Node temp = new Node(); temp.data = data; temp.left = temp.right = null; return temp;}// Driver Codepublic 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 |
Javascript
<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.
C++
// CPP program to print minimum element// in each level of binary tree.#include <iostream>#include <queue>#include <vector>#define INT_MAX 10e6using namespace std;// A Binary Tree Nodestruct Node { int data; struct Node *left, *right;};// return height of treeint 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 Treevoid 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 nodeNode* newNode(int data){ Node* temp = new Node; temp->data = data; temp->left = temp->right = NULL; return temp;}// Driver program to test above functionsint 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
// JAVA program to print minimum element// in each level of binary tree.import java.util.*;class GFG{// A Binary Tree Nodestatic class Node { int data; Node left, right;};// return height of treestatic 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 Treestatic 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 nodestatic Node newNode(int data){ Node temp = new Node(); temp.data = data; temp.left = temp.right = null; return temp;}// Driver codepublic 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
# Python3 program to print minimum element # in each level of binary tree. # Importing Queuefrom 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 pdef 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 Codeif __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#
// C# program to print minimum element// in each level of binary tree.using System;using System.Collections.Generic;class GFG{// A Binary Tree Nodeclass Node { public int data; public Node left, right;};// return height of treestatic 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 Treestatic 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 nodestatic Node newNode(int data){ Node temp = new Node(); temp.data = data; temp.left = temp.right = null; return temp;}// Driver codepublic 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 |
Javascript
<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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