# 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++ implementation to print largest ` `// value in each level of Binary Tree ` `#include ` ` `  `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 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 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 q = ``new` `LinkedList(); ` `        ``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); ` `    ``} ` `}`

 `# 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 ` `        ``4  ` `        ``/ \  ` `        ``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# 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 q = ``new` `Queue(); ` `        ``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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

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 :