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 the n elements divided by n and rounded down to the nearest integer.

• A subtree of root is a tree consisting of root 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：

1. 后序遍历：从根节点开始，递归地遍历左子树和右子树。

2. 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。

3. 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