Average of Levels in Binary Tree

#leetcode #dsa #theabbie

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.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def averageOfLevels(self, root: Optional[TreeNode]) -> List[float]:
        if not root:
            return []
        queue = [(root, 0)]
        levels = {}
        level = 0
        while len(queue) > 0:
            curr, level = queue.pop(0)
            levels[level] = levels.get(level, []) + [curr.val]
            if curr.left:
                queue.append((curr.left, level + 1))
            if curr.right:
                queue.append((curr.right, level + 1))
        return [sum(levels[l])/len(levels[l]) for l in range(level + 1)]