Average of Levels in Binary Tree

Problem

Given the root of a binary tree, return the average value of the nodes on each level in the form of an array. Answers within 10-5 of the actual answer will be accepted.

 

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [3.00000,14.50000,11.00000]
Explanation: The average value of nodes on level 0 is 3, on level 1 is 14.5, and on level 2 is 11.
Hence return [3, 14.5, 11].

Example 2:

Input: root = [3,9,20,15,7]
Output: [3.00000,14.50000,11.00000]

 

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• -231 <= Node.val <= 231 - 1

Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var averageOfLevels = function(root) {
    const queue = [root];
    const averages = [];

    // level order traversal using BFS
    while (queue.length) {
        let average = 0;
        const count = queue.length;

        for (let i = 0; i < count; i++) {
            const node = queue.shift();
            average += node.val;

            if (node.left) {
                queue.push(node.left);
            }
            if (node.right) {
                queue.push(node.right);
            }
        }
        averages.push(average / count);
    }
    return averages;
};

We will use BFS to perform a level order traversal on the tree root and get the average value of the nodes on each level.