Smallest value in each level of Binary Tree

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1: Using In-order traversal
Approach:- The idea is to recursively traverse tree in a in-order fashion. Root is considered to be at zeroth level. First find the height of tree and store it into res. res array store every smallest element in each level of binary tree.

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

C++

 // CPP program to print smallest element // in each level of binary tree. #include #include #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); }    // Inorder Traversal // Search minimum element in each level and  // store it into vector array. void printPerLevelMinimum(Node* root,                    vector& 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 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

 // 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

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 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

Output:

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

Method 2: Using level order Traversal
Approach:- The idea is to perform iterative level order traversal of the binary tree using 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 #include #include #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 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; }

Python3

 # Python3 program to prminimum 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 get of queue          node = q.queue          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

Output:

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

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