Description
https://leetcode.com/problems/average-of-levels-in-binary-tree/
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,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
Explanation
Do breadth-first search and calculate average on levels.
Python 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: TreeNode) -> List[float]:
results = []
queue = []
queue.append(root)
while queue:
size = len(queue)
level = []
for i in range(size):
node = queue.pop(0)
level.append(node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
results.append(sum(level) / len(level))
return results- Time Complexity: O(N).
- Space Complexity: O(N).