LeetCode #2265 — MEDIUM

Count Nodes Equal to Average of Subtree

Move from brute-force thinking to an efficient approach using tree strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Tree

Example 1

[4,8,5,0,1,null,6]

Example 2

[1]

Core Insight

What unlocks the optimal approach

What information do we need to calculate the average? We need the sum of the values and the number of values.
Create a recursive function that returns the size of a node’s subtree, and the sum of the values of its subtree.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2265: Count Nodes Equal to Average of Subtree
/**
 * 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) {
        dfs(root);
        return ans;
    }

    private int[] dfs(TreeNode root) {
        if (root == null) {
            return new int[2];
        }
        var l = dfs(root.left);
        var 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};
    }
}

// Accepted solution for LeetCode #2265: Count Nodes Equal to Average of Subtree
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func averageOfSubtree(root *TreeNode) (ans int) {
	var dfs func(root *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, n := ls+rs+root.Val, ln+rn+1
		if s/n == root.Val {
			ans++
		}
		return s, n
	}
	dfs(root)
	return
}

# Accepted solution for LeetCode #2265: Count Nodes Equal to Average of Subtree
class Solution:
    def averageOfSubtree(self, root: TreeNode) -> int:
        def dfs(root) -> tuple:
            if not root:
                return 0, 0
            ls, ln = dfs(root.left)
            rs, rn = dfs(root.right)
            s = ls + rs + root.val
            n = ln + rn + 1
            nonlocal ans
            ans += int(s // n == root.val)
            return s, n

        ans = 0
        dfs(root)
        return ans

// Accepted solution for LeetCode #2265: Count Nodes Equal to Average of Subtree
/**
 * [2265] Count Nodes Equal to Average of Subtree
 *
 * 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:
 * <img src="https://assets.leetcode.com/uploads/2022/03/15/image-20220315203925-1.png" style="width: 300px; height: 212px;" />
 * 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:
 * <img src="https://assets.leetcode.com/uploads/2022/03/26/image-20220326133920-1.png" style="width: 80px; height: 76px;" />
 * 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
 *
 */
pub struct Solution {}
use crate::util::tree::{TreeNode, to_tree};

// problem: https://leetcode.com/problems/count-nodes-equal-to-average-of-subtree/
// discuss: https://leetcode.com/problems/count-nodes-equal-to-average-of-subtree/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    pub fn average_of_subtree(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        let mut result = 0;
        Self::dfs_helper(root, &mut result);
        result
    }

    fn dfs_helper(root: Option<Rc<RefCell<TreeNode>>>, counter: &mut i32) -> (i32, i32) {
        if let Some(root) = root {
            let root = root.borrow();
            let (lsum, lcount) = Self::dfs_helper(root.left.clone(), counter);
            let (rsum, rcount) = Self::dfs_helper(root.right.clone(), counter);
            let total_sum = root.val + lsum + rsum;
            let total_count = 1 + lcount + rcount;
            if root.val == total_sum / total_count {
                *counter += 1;
            }
            (total_sum, total_count)
        } else {
            (0, 0)
        }
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2265_example_1() {
        let root = tree![4, 8, 5, 0, 1, null, 6];

        let result = 5;

        assert_eq!(Solution::average_of_subtree(root), result);
    }

    #[test]
    fn test_2265_example_2() {
        let root = tree![1];

        let result = 1;

        assert_eq!(Solution::average_of_subtree(root), result);
    }
}

// Accepted solution for LeetCode #2265: Count Nodes Equal to Average of Subtree
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function averageOfSubtree(root: TreeNode | null): number {
    let ans: number = 0;
    const dfs = (root: TreeNode | null): [number, number] => {
        if (!root) {
            return [0, 0];
        }
        const [ls, ln] = dfs(root.left);
        const [rs, rn] = dfs(root.right);
        const s = ls + rs + root.val;
        const n = ln + rn + 1;
        if (Math.floor(s / n) === root.val) {
            ++ans;
        }
        return [s, n];
    };
    dfs(root);
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.