081.md

637. Average of Levels in Binary Tree

Difficulty: 🟢 Easy

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.

Examples:

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

Solutions

O(n) solution

Python3

# 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]:
        result, level = [], [root]
        
        while level:
            result.append(sum([node.val for node in level])/len(level))
            level = [node.left for node in level if node.left] + [node.right for node in level if node.right]
        
        return result

The given solution provides an iterative approach to find the average value of nodes on each level of a binary tree.

Here is a step-by-step overview of the solution:

1. Initialize an empty list result to store the average values of each level.
2. Initialize a list level with the root node.
3. Enter a loop that continues until there are no nodes left in the level:
   • Calculate the average value of the nodes in the current level by summing up the values of the nodes and dividing by the number of nodes.
   • Append the average value to the result.
   • Create a new list next_level to store the nodes of the next level.
   • Iterate through the nodes in the current level and add their left and right children (if they exist) to the next_level.
4. Set the level to be the next_level for the next iteration.
5. Return the result, which contains the average values of each level.

Complexity Analysis

The time complexity for this solution is O(n), where n is the number of nodes in the binary tree. We visit each node once.

The space complexity is O(m), where m is the maximum number of nodes at any level. In the worst case, the maximum number of nodes at any level can be n/2, so the space complexity is O(n/2), which simplifies to O(n).

Summary

The given solution provides an iterative approach to find the average value of nodes on each level of a binary tree. It uses a queue-like approach to traverse the tree level by level and calculate the average values. The solution has a time complexity of O(n) and a space complexity of O(m), where n is the number of nodes and m is the maximum number of nodes at any level in the binary tree.

NB: If you want to get community points please suggest solutions in other languages as merge requests.