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Formatted question description: https://leetcode.ca/all/2265.html

# 2265. Count Nodes Equal to Average of Subtree

• Difficulty: Medium.
• Related Topics: Tree, Depth-First Search, Binary Tree.
• Similar Questions: Maximum Average Subtree, Insufficient Nodes in Root to Leaf Paths, Count Nodes Equal to Sum of Descendants.

## Problem

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

## Solution (Java, C++, Python)

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
private int ans = 0;

public int averageOfSubtree(TreeNode root) {
dfs(root);
return ans;
}

private int[] dfs(TreeNode node) {
if (node == null) {
return new int[] {0, 0};
}
int[] left = dfs(node.left);
int[] right = dfs(node.right);
int currsum = left[0] + right[0] + node.val;
int currcount = left[1] + right[1] + 1;
if (currsum / currcount == node.val) {
++ans;
}
return new int[] {currsum, currcount};
}
}

############

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
private int ans;

public int averageOfSubtree(TreeNode root) {
ans = 0;
dfs(root);
return ans;
}

private int[] dfs(TreeNode root) {
if (root == null) {
return new int[] {0, 0};
}
int[] l = dfs(root.left);
int[] r = dfs(root.right);
int s = l[0] + r[0] + root.val;
int n = l[1] + r[1] + 1;
if (s / n == root.val) {
++ans;
}
return new int[] {s, n};
}
}

• # 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: Optional[TreeNode]) -> int:
def dfs(root):
if root is None:
return 0, 0
ls, ln = dfs(root.left)
rs, rn = dfs(root.right)
s = ls + rs + root.val
n = ln + rn + 1
if s // n == root.val:
nonlocal ans
ans += 1
return s, n

ans = 0
dfs(root)
return ans

############

# 2265. Count Nodes Equal to Average of Subtree
# https://leetcode.com/problems/count-nodes-equal-to-average-of-subtree/

# 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: Optional[TreeNode]) -> int:
res = 0

def go(node):
if not node: return (0, 0)

lCount, lValues = go(node.left)
rCount, rValues = go(node.right)

nodeCount, nodeValues = 1 + lCount + rCount, node.val + lValues + rValues

average = nodeValues // nodeCount

if node.val == average:
nonlocal res

res += 1

return (nodeCount, nodeValues)

go(root)

return res


• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int ans;
int averageOfSubtree(TreeNode* root) {
ans = 0;
dfs(root);
return ans;
}

vector<int> dfs(TreeNode* root) {
if (!root) return {0, 0};
auto l = dfs(root->left);
auto r = dfs(root->right);
int s = l[0] + r[0] + root->val;
int n = l[1] + r[1] + 1;
if (s / n == root->val) ++ans;
return {s, n};
}
};

• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func averageOfSubtree(root *TreeNode) int {
ans := 0
var dfs func(*TreeNode) (int, int)
dfs = func(root *TreeNode) (int, int) {
if root == nil {
return 0, 0
}
ls, ln := dfs(root.Left)
rs, rn := dfs(root.Right)
s := ls + rs + root.Val
n := ln + rn + 1
if s/n == root.Val {
ans++
}
return s, n
}
dfs(root)
return ans
}


Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).