Averages of Levels in Binary Tree

• Difficulty Level : Easy
• Last Updated : 01 Jul, 2021

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].

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 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 codeint 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 toleft child and a pointer to right child */static class Node {    int val;    Node left, right;} /* Function to print the average value of thenodes 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 anew node with the given data andNULL 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 codepublic 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 Queuefrom 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 leveldef 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            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 codeif __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

Javascript



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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