# 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
```

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

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 ` ` `  `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 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; ` `} `

## 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 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 program to test above  ` `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``);  ` ` `  `    ``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 ` `        ``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``) ` `    ``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 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);  ` ` `  `    ``largestValueInEachLevel(root);  ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```4 9 7
```

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

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