Given the root
of a binary tree, return the number of nodes where the value of the node is equal to the average of the values in its subtree.
Note:
- The average of
n
elements is the sum of then
elements divided byn
and rounded down to the nearest integer. - A subtree of
root
is a tree consisting ofroot
and all of its descendants.
Example 1:
Input: root = [4,8,5,0,1,null,6]
Output: 5
Explanation:
For the node with value 4: The average of its subtree is (4 + 8 + 5 + 0 + 1 + 6) / 6 = 24 / 6 = 4.
For the node with value 5: The average of its subtree is (5 + 6) / 2 = 11 / 2 = 5.
For the node with value 0: The average of its subtree is 0 / 1 = 0.
For the node with value 1: The average of its subtree is 1 / 1 = 1.
For the node with value 6: The average of its subtree is 6 / 1 = 6.
Example 2:
Input: root = [1]
Output: 1
Explanation: For the node with value 1: The average of its subtree is 1 / 1 = 1.
Constraints:
- The number of nodes in the tree is in the range
[1, 1000]
. -
0 <= Node.val <= 1000
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 sumcount(self, root):
total = 0
n = 0
if root:
ltotal, ln = self.sumcount(root.left)
rtotal, rn = self.sumcount(root.right)
total = ltotal + rtotal + root.val
n = ln + rn + 1
if root.val == total // n:
self.ctr += 1
return (total, n)
def averageOfSubtree(self, root: Optional[TreeNode]) -> int:
self.ctr = 0
self.sumcount(root)
return self.ctr
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