Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.
Example 1:
Input:
3
/ \
9 20
/ \
15 7
Output: [3, 14.5, 11]
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].
Note: The range of node’s value is in the range of 32-bit signed integer.
Solution in python:
-
BFS
# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: def averageOfLevels(self, root: TreeNode) -> List[float]: ave = [] queue = [root] num = 1 values, new_num= 0, 0 while queue: for i in range(num): node = queue.pop(0) values += node.val if node.left: queue.append(node.left) new_num += 1 if node.right: queue.append(node.right) new_num += 1 ave.append(values/num) num = new_num new_num, values = 0, 0 return ave
Complexity analysis:
Time complexity: O(n)
. The whole tree is traversed atmost once. Here, n refers to the number of nodes in the given binary tree.
Space complexity: O(m)
. The size of queue or ave can grow upto atmost the maximum number of nodes at any level in the given binary tree. Here, m refers to the maximum mumber of nodes at any level in the input tree.
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