Largest value in each level of Binary Tree | Set-2 (Iterative Approach)

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 program to test above
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);
  
    largestValueInEachLevel(root);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 program to test above 
public static void main(String[] args) 
    /* Construct a Binary Tree 
        
    / \ 
    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); 
  
    largestValueInEachLevel(root); 
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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)
    largestValueInEachLevel(root)
  
# This code is contributed
# Shubham Singh(SHUBHAMSINGH10)

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 
        
    / \ 
    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); 
  
    largestValueInEachLevel(root); 
}
}
  
// This code is contributed by PrinciRaj1992

chevron_right



Output:

4 9 7

Time Complexity: O(n).
Auxiliary Space: O(n).



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.





Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.