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Largest value in each level of Binary Tree | Set-2 (Iterative Approach)
  • Difficulty Level : Easy
  • Last Updated : 22 Oct, 2020

Given a binary tree containing n nodes. The problem is to find and print the largest value present in each level.
Examples: 

Input :
        1
       / \
      2   3 
Output : 1 3

Input : 
        4
       / \
      9   2
     / \   \
    3   5   7 
Output : 4 9 7

Approach: In the previous post, a recursive method have been discussed. In this post an iterative method has been discussed. The idea is to perform iterative level order traversal of the binary tree using queue. While traversing keep max variable which stores the maximum element of the current level of the tree being processed. When the level is completely traversed, print that max value.  

C++




// C++ implementation to print largest
// value in each level of Binary Tree
#include <bits/stdc++.h>
  
using namespace std;
  
// structure of a node of binary tree
struct Node {
    int data;
    Node *left, *right;
};
  
// function to get a new node
Node* newNode(int data)
{
    // allocate space
    Node* temp = new Node;
  
    // put in the data
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
  
// function to print largest value
// in each level of Binary Tree
void largestValueInEachLevel(Node* root)
{
    // if tree is empty
    if (!root)
        return;
  
    queue<Node*> q;
    int nc, max;
  
    // push root to the queue 'q'
    q.push(root);
  
    while (1) {
        // node count for the current level
        nc = q.size();
  
        // if true then all the nodes of
        // the tree have been traversed
        if (nc == 0)
            break;
  
        // maximum element for the current
        // level
        max = INT_MIN;
  
        while (nc--) {
  
            // get the front element from 'q'
            Node* front = q.front();
  
            // remove front element from 'q'
            q.pop();
  
            // if true, then update 'max'
            if (max < front->data)
                max = front->data;
  
            // if left child exists
            if (front->left)
                q.push(front->left);
  
            // if right child exists
            if (front->right)
                q.push(front->right);
        }
  
        // print maximum element of
        // current level
        cout << max << " ";
    }
}
  
// Driver code
int main()
{
    /* Construct a Binary Tree
        4
       / \
      9   2
     / \   \
    3   5   7 */
  
    Node* root = NULL;
    root = newNode(4);
    root->left = newNode(9);
    root->right = newNode(2);
    root->left->left = newNode(3);
    root->left->right = newNode(5);
    root->right->right = newNode(7);
  
    // Function call
    largestValueInEachLevel(root);
  
    return 0;
}


Java




// Java implementation to print largest
// value in each level of Binary Tree
import java.util.*;
class GfG {
  
    // structure of a node of binary tree
    static class Node 
    {
        int data;
        Node left = null;
        Node right = null;
    }
  
    // function to get a new node
    static Node newNode(int val)
    {
        // allocate space
        Node temp = new Node();
  
        // put in the data
        temp.data = val;
        temp.left = null;
        temp.right = null;
        return temp;
    }
  
    // function to print largest value
    // in each level of Binary Tree
    static void largestValueInEachLevel(Node root)
    {
        // if tree is empty
        if (root == null)
            return;
  
        Queue<Node> q = new LinkedList<Node>();
        int nc, max;
  
        // push root to the queue 'q'
        q.add(root);
  
        while (true
        {
            // node count for the current level
            nc = q.size();
  
            // if true then all the nodes of
            // the tree have been traversed
            if (nc == 0)
                break;
  
            // maximum element for the current
            // level
            max = Integer.MIN_VALUE;
  
            while (nc != 0
            {
  
                // get the front element from 'q'
                Node front = q.peek();
  
                // remove front element from 'q'
                q.remove();
  
                // if true, then update 'max'
                if (max < front.data)
                    max = front.data;
  
                // if left child exists
                if (front.left != null)
                    q.add(front.left);
  
                // if right child exists
                if (front.right != null)
                    q.add(front.right);
                nc--;
            }
  
            // print maximum element of
            // current level
            System.out.println(max + " ");
        }
    }
  
    // Driver code
    public static void main(String[] args)
    {
        /* Construct a Binary Tree
            4
        / \
        9 2
        / \ \
        3 5 7 */
  
        Node root = null;
        root = newNode(4);
        root.left = newNode(9);
        root.right = newNode(2);
        root.left.left = newNode(3);
        root.left.right = newNode(5);
        root.right.right = newNode(7);
  
        // Function call
        largestValueInEachLevel(root);
    }
}


Python3




# Python program to print largest value
# on each level of binary tree
  
INT_MIN = -2147483648
  
# Helper function that allocates a new
# node with the given data and None left
# and right pointers.
  
  
class newNode:
  
    # Constructor to create a new node
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
  
# function to find largest values
  
  
def largestValueInEachLevel(root):
    if (not root):
        return
    q = []
    nc = 10
    max = 0
    q.append(root)
    while (1):
        # node count for the current level
        nc = len(q)
  
        # if true then all the nodes of
        # the tree have been traversed
        if (nc == 0):
            break
  
        # maximum element for the current
        # level
        max = INT_MIN
        while (nc):
  
            # get the front element from 'q'
            front = q[0]
  
            # remove front element from 'q'
            q = q[1:]
  
            # if true, then update 'max'
            if (max < front.data):
                max = front.data
  
            # if left child exists
            if (front.left):
                q.append(front.left)
  
            # if right child exists
            if (front.right != None):
                q.append(front.right)
            nc -= 1
  
        # print maximum element of
        # current level
        print(max, end=" ")
  
  
# Driver Code
if __name__ == '__main__':
    """ Let us construct the following Tree
        
        / \ 
        9 2 
    / \ \
    3 5 7 """
    root = newNode(4)
    root.left = newNode(9)
    root.right = newNode(2)
    root.left.left = newNode(3)
    root.left.right = newNode(5)
    root.right.right = newNode(7)
      
    # Function call
    largestValueInEachLevel(root)
  
# This code is contributed
# Shubham Singh(SHUBHAMSINGH10)


C#




// C# implementation to print largest
// value in each level of Binary Tree
using System;
using System.Collections.Generic;
  
class GfG
{
  
    // structure of a node of binary tree
    class Node 
    {
        public int data;
        public Node left = null;
        public Node right = null;
    }
  
    // function to get a new node
    static Node newNode(int val)
    {
        // allocate space
        Node temp = new Node();
  
        // put in the data
        temp.data = val;
        temp.left = null;
        temp.right = null;
        return temp;
    }
  
    // function to print largest value
    // in each level of Binary Tree
    static void largestValueInEachLevel(Node root)
    {
        // if tree is empty
        if (root == null)
            return;
  
        Queue<Node> q = new Queue<Node>();
        int nc, max;
  
        // push root to the queue 'q'
        q.Enqueue(root);
  
        while (true
        {
            // node count for the current level
            nc = q.Count;
  
            // if true then all the nodes of
            // the tree have been traversed
            if (nc == 0)
                break;
  
            // maximum element for the current
            // level
            max = int.MinValue;
  
            while (nc != 0) 
            {
                // get the front element from 'q'
                Node front = q.Peek();
  
                // remove front element from 'q'
                q.Dequeue();
  
                // if true, then update 'max'
                if (max < front.data)
                    max = front.data;
  
                // if left child exists
                if (front.left != null)
                    q.Enqueue(front.left);
  
                // if right child exists
                if (front.right != null)
                    q.Enqueue(front.right);
                nc--;
            }
  
            // print maximum element of
            // current level
            Console.Write(max + " ");
        }
    }
  
    // Driver code
    public static void Main(String[] args)
    {
        /* Construct a Binary Tree
            4
        / \
        9 2
        / \ \
        3 5 7 */
  
        Node root = null;
        root = newNode(4);
        root.left = newNode(9);
        root.right = newNode(2);
        root.left.left = newNode(3);
        root.left.right = newNode(5);
        root.right.right = newNode(7);
  
        // Function call
        largestValueInEachLevel(root);
    }
}
  
// This code is contributed by PrinciRaj1992


Output

4 9 7 

Complexity Analysis:

Time Complexity : O(N) where N is the total number of nodes in the tree.



In level order traversal, every node of the tree is processed once and hence the complexity due to the level order traversal is O(N) if there are total N nodes in the tree. Also, while processing every node, we are maintaining the maximum element at each level, however, this does not affect the overall time complexity. Therefore, the time complexity is O(N).

Auxiliary Space : O(w) where w is the maximum width of the tree.

In level order traversal, a queue is maintained whose maximum size at any moment can go upto the maximum width of the binary tree.
 

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