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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).

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