2265. Count Nodes Equal to Average of Subtree
tree dfs
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
分析:
这道题目要求我们统计二叉树中满足以下条件的节点数目:节点的值等于其子树(包括该节点自身)所有节点值的平均值。为了高效地解决这个问题,可以采用后序遍历的方式,因为后序遍历会先处理子节点,再处理父节点,这样在处理每个节点时,我们已经知道了其左右子树的信息。
具体步骤如下:
后序遍历:从根节点开始,递归地遍历左子树和右子树。
计算子树和与节点数:对于每个节点,计算其左右子树的和与节点数目,然后加上当前节点的值,得到整个子树的和与节点数目。
判断条件:检查当前节点的值是否等于子树和的平均值(整数除法),如果满足条件,则增加计数器。
# 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 averageOfSubtree(self, root: TreeNode) -> int:
res = 0
def dfs(root):
nonlocal res
if not root:
return (0,0)
lv, lcnt = dfs(root.left)
rv,rcnt = dfs(root.right)
val = root.val + lv + rv
cnt = lcnt + rcnt + 1
if val // cnt == root.val:
res += 1
return (val, cnt)
dfs(root)
return res
Last updated
Was this helpful?