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)]
Top comments (0)