# Averages of Levels in Binary Tree

Given a non-empty binary tree, print the average value of the nodes on each level.

Examples:

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

Output : [4 5.5 5]
The average value of nodes on level 0 is 4,
on level 1 is 5.5, and on level 2 is 5.
Hence, print [4 5.5 5].
```

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

The idea is based on Level order traversal line by line | Set 2 (Using Two Queues)

1. Start by pushing the root node into the queue. Then, remove a node from the front of the queue.
2. For every node removed from the queue, push all its children into a new temporary queue.
3. Keep on popping nodes from the queue and adding these node’ s children to the temporary queue till queue becomes empty.
4. Every time queue becomes empty, it indicates that one level of the tree has been considered.
5. While pushing the nodes into temporary queue, keep a track of the sum of the nodes along with the number of nodes pushed and find out the average of the nodes on each level by making use of these sum and count values.
6. After each level has been considered, again initialize the queue with temporary queue and continue the process till both queues become empty.

## C++

 `// C++ program to find averages of all levels ` `// in a binary tree. ` `#include ` `using` `namespace` `std; ` ` `  `/* A binary tree node has data, pointer to ` `   ``left child and a pointer to right child */` `struct` `Node { ` `    ``int` `val; ` `    ``struct` `Node* left, *right; ` `}; ` ` `  `/* Function to print the average value of the ` `   ``nodes on each level */` `void` `averageOfLevels(Node* root) ` `{ ` `    ``vector<``float``> res; ` ` `  `    ``// Traversing level by level ` `    ``queue q; ` `    ``q.push(root); ` ` `  `    ``while` `(!q.empty()) { ` ` `  `        ``// Compute sum of nodes and ` `        ``// count of nodes in current ` `        ``// level. ` `        ``int` `sum = 0, count = 0; ` `        ``queue temp; ` `        ``while` `(!q.empty()) { ` `            ``Node* n = q.front(); ` `            ``q.pop(); ` `            ``sum += n->val; ` `            ``count++; ` `            ``if` `(n->left != NULL) ` `                ``temp.push(n->left); ` `            ``if` `(n->right != NULL) ` `                ``temp.push(n->right); ` `        ``} ` `        ``q = temp; ` `        ``cout << (sum * 1.0 / count) << ``" "``; ` `    ``} ` `} ` ` `  `/* Helper function that allocates a ` `   ``new node with the given data and ` `   ``NULL left and right pointers. */` `Node* newNode(``int` `data) ` `{ ` `    ``Node* temp = ``new` `Node; ` `    ``temp->val = data; ` `    ``temp->left = temp->right = NULL; ` `    ``return` `temp; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``/* Let us construct a Binary Tree ` `        ``4 ` `       ``/ \ ` `      ``2   9 ` `     ``/ \   \ ` `    ``3   5   7 */` ` `  `    ``Node* root = NULL; ` `    ``root = newNode(4); ` `    ``root->left = newNode(2); ` `    ``root->right = newNode(9); ` `    ``root->left->left = newNode(3); ` `    ``root->left->right = newNode(8); ` `    ``root->right->right = newNode(7); ` `    ``averageOfLevels(root); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find averages of all levels  ` `// in a binary tree.  ` `import` `java.util.*; ` `class` `GfG { ` ` `  `/* A binary tree node has data, pointer to  ` `left child and a pointer to right child */` `static` `class` `Node {  ` `    ``int` `val;  ` `    ``Node left, right;  ` `} ` ` `  `/* Function to print the average value of the  ` `nodes on each level */` `static` `void` `averageOfLevels(Node root)  ` `{  ` `    ``//vector res;  ` ` `  `    ``// Traversing level by level  ` `    ``Queue q = ``new` `LinkedList ();  ` `    ``q.add(root);  ` `    ``int` `sum = ``0``, count  = ``0``; ` ` `  `    ``while` `(!q.isEmpty()) {  ` ` `  `        ``// Compute sum of nodes and  ` `        ``// count of nodes in current  ` `        ``// level.  ` `        ``sum = ``0``; ` `        ``count = ``0``;  ` `        ``Queue temp = ``new` `LinkedList ();  ` `        ``while` `(!q.isEmpty()) {  ` `            ``Node n = q.peek();  ` `            ``q.remove();  ` `            ``sum += n.val;  ` `            ``count++;  ` `            ``if` `(n.left != ``null``)  ` `                ``temp.add(n.left);  ` `            ``if` `(n.right != ``null``)  ` `                ``temp.add(n.right);  ` `        ``}  ` `        ``q = temp;  ` `        ``System.out.print((sum * ``1.0` `/ count) + ``" "``);  ` `    ``}  ` `}  ` ` `  `/* Helper function that allocates a  ` `new node with the given data and  ` `NULL left and right pointers. */` `static` `Node newNode(``int` `data)  ` `{  ` `    ``Node temp = ``new` `Node();  ` `    ``temp.val = data;  ` `    ``temp.left = ``null``; ` `    ``temp.right = ``null``;  ` `    ``return` `temp;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String[] args)  ` `{  ` `    ``/* Let us construct a Binary Tree  ` `        ``4  ` `    ``/ \  ` `    ``2 9  ` `    ``/ \ \  ` `    ``3 5 7 */` ` `  `    ``Node root = ``null``;  ` `    ``root = newNode(``4``);  ` `    ``root.left = newNode(``2``);  ` `    ``root.right = newNode(``9``);  ` `    ``root.left.left = newNode(``3``);  ` `    ``root.left.right = newNode(``5``);  ` `    ``root.right.right = newNode(``7``);  ` `    ``System.out.println(``"Averages of levels : "``); ` `    ``System.out.print(``"["``); ` `    ``averageOfLevels(root); ` `    ``System.out.println(``"]"``); ` `} ` `}  `

## Python3

 `# Python3 program to find averages of  ` `# all levels in a binary tree.  ` ` `  `# Importing Queue ` `from` `queue ``import` `Queue ` ` `  `# Helper class that allocates a  ` `# new node with the given data and  ` `# None left and right pointers.  ` `class` `newNode: ` `    ``def` `__init__(``self``, data): ` `        ``self``.val ``=` `data  ` `        ``self``.left ``=` `self``.right ``=` `None` `     `  `# Function to print the average value  ` `# of the nodes on each level  ` `def` `averageOfLevels(root): ` ` `  `    ``# Traversing level by level  ` `    ``q ``=` `Queue() ` `    ``q.put(root)  ` `    ``while` `(``not` `q.empty()): ` ` `  `        ``# Compute Sum of nodes and  ` `        ``# count of nodes in current  ` `        ``# level.  ` `        ``Sum` `=` `0` `        ``count ``=` `0` `        ``temp ``=` `Queue()  ` `        ``while` `(``not` `q.empty()):  ` `            ``n ``=` `q.queue[``0``]  ` `            ``q.get()  ` `            ``Sum` `+``=` `n.val  ` `            ``count ``+``=` `1` `            ``if` `(n.left !``=` `None``): ` `                ``temp.put(n.left)  ` `            ``if` `(n.right !``=` `None``):  ` `                ``temp.put(n.right) ` `        ``q ``=` `temp  ` `        ``print``((``Sum` `*` `1.0` `/` `count), end ``=` `" "``) ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `'__main__'``: ` ` `  `    ``# Let us construct a Binary Tree  ` `    ``#     4  ` `    ``# / \  ` `    ``# 2 9  ` `    ``# / \ \  ` `    ``# 3 5 7  ` `    ``root ``=` `None` `    ``root ``=` `newNode(``4``)  ` `    ``root.left ``=` `newNode(``2``)  ` `    ``root.right ``=` `newNode(``9``)  ` `    ``root.left.left ``=` `newNode(``3``)  ` `    ``root.left.right ``=` `newNode(``8``)  ` `    ``root.right.right ``=` `newNode(``7``)  ` `    ``averageOfLevels(root) ` ` `  `# This code is contributed by PranchalK `

## C#

 `// C# program to find averages of all levels  ` `// in a binary tree. ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GfG  ` `{ ` ` `  `    ``/* A binary tree node has data, pointer to  ` `    ``left child and a pointer to right child */` `    ``class` `Node  ` `    ``{  ` `        ``public` `int` `val;  ` `        ``public` `Node left, right;  ` `    ``} ` ` `  `    ``/* Function to print the average value of the  ` `    ``nodes on each level */` `    ``static` `void` `averageOfLevels(Node root)  ` `    ``{  ` `        ``//vector res;  ` ` `  `        ``// Traversing level by level  ` `        ``Queue q = ``new` `Queue ();  ` `        ``q.Enqueue(root);  ` `        ``int` `sum = 0, count = 0; ` ` `  `        ``while` `((q.Count!=0))  ` `        ``{  ` ` `  `            ``// Compute sum of nodes and  ` `            ``// count of nodes in current  ` `            ``// level.  ` `            ``sum = 0; ` `            ``count = 0;  ` `            ``Queue temp = ``new` `Queue ();  ` `            ``while` `(q.Count != 0)  ` `            ``{  ` `                ``Node n = q.Peek();  ` `                ``q.Dequeue();  ` `                ``sum += n.val;  ` `                ``count++;  ` `                ``if` `(n.left != ``null``)  ` `                    ``temp.Enqueue(n.left);  ` `                ``if` `(n.right != ``null``)  ` `                    ``temp.Enqueue(n.right);  ` `            ``}  ` `            ``q = temp;  ` `            ``Console.Write((sum * 1.0 / count) + ``" "``);  ` `        ``}  ` `    ``}  ` ` `  `    ``/* Helper function that allocates a  ` `    ``new node with the given data and  ` `    ``NULL left and right pointers. */` `    ``static` `Node newNode(``int` `data)  ` `    ``{  ` `        ``Node temp = ``new` `Node();  ` `        ``temp.val = data;  ` `        ``temp.left = ``null``; ` `        ``temp.right = ``null``;  ` `        ``return` `temp;  ` `    ``}  ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{  ` `        ``/* Let us construct a Binary Tree  ` `            ``4  ` `        ``/ \  ` `        ``2 9  ` `        ``/ \ \  ` `        ``3 5 7 */` ` `  `        ``Node root = ``null``;  ` `        ``root = newNode(4);  ` `        ``root.left = newNode(2);  ` `        ``root.right = newNode(9);  ` `        ``root.left.left = newNode(3);  ` `        ``root.left.right = newNode(5);  ` `        ``root.right.right = newNode(7);  ` `        ``Console.WriteLine(``"Averages of levels : "``); ` `        ``Console.Write(``"["``); ` `        ``averageOfLevels(root); ` `        ``Console.WriteLine(``"]"``); ` `    ``} ` `}  ` ` `  `// This code has been contributed by ` `// 29AjayKumar `

Output:

```Average of levels:
[4 5.5 5]
```

Complexity Analysis:

• Time complexity : O(n).
The whole tree is traversed atmost once. Here, n refers to the number of nodes in the given binary tree.
• Auxiliary Space : O(n).
The size of queues can grow upto atmost the maximum number of nodes at any level in the given binary tree. Here, n refers to the maximum number of nodes at any level in the input tree.

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