Subtree with Maximum Average- Amazon Top Interview Questions

Problem Statement :

Given a binary tree root, return the maximum average value of a subtree. A subtree is defined to be some node in root including all of its descendants. A subtree average is the sum of the node values divided by the number of nodes.


1 ≤ n ≤ 100,000 where n is the number of nodes in root

Example 1


root = [1, [3, null, null], [7, [4, null, null], null]]




The subtree rooted at 7 has the highest average with (7 + 4) / 2.

Example 2


root = [1, [2, null, null], [3, [4, null, null], null]]




The subtree rooted at 4 has the highest average with 4 / 1.

Solution :


                        Solution in C++ :

void maxAvg(Tree* root, int& c, int& sum, double& big) {
    if (!root) return;
    sum += root->val;
    c += 1;
    int lCount = 0, rCount = 0, lsum = 0, rsum = 0;
    maxAvg(root->left, lCount, lsum, big);
    maxAvg(root->right, rCount, rsum, big);
    c += lCount + rCount;
    sum += lsum + rsum;
    double avg = (double)sum / (1.0 * c);
    if (avg > big) big = avg;
double solve(Tree* root) {
    if (!root) return 0.0;
    double res = 0;
    int c = 0;
    int sum = 0;
    maxAvg(root, c, sum, res);
    return res;

                        Solution in Java :

import java.util.*;

 * public class Tree {
 *   int val;
 *   Tree left;
 *   Tree right;
 * }
class Solution {
    public double solve(Tree root) {
        return rec(root).maxAvg;

    public ReturnType rec(Tree node) {
        if (node == null)
            return new ReturnType(0, 0, 0);
        ReturnType l = rec(node.left);
        ReturnType r = rec(node.right);
        int curSum = (l.sum + r.sum + node.val);
        int curCnt = (l.cnt + r.cnt + 1);
        double curAvg = (double) curSum / curCnt;
        double maxAvg = Math.max(Math.max(l.maxAvg, r.maxAvg), curAvg);
        return new ReturnType(curCnt, curSum, maxAvg);

    public class ReturnType {
        int cnt;
        int sum;
        double maxAvg;

        public ReturnType(int cnt, int sum, double maxAvg) {
            this.cnt = cnt;
            this.sum = sum;
            this.maxAvg = maxAvg;

                        Solution in Python : 
class Solution:
    def solve(self, root):

        ans = 0

        def dfs(root):
            nonlocal ans
            if not root:
                return (0, 0)

            left_sum, nl = dfs(root.left)
            right_sum, nr = dfs(root.right)

            cur_sum = left_sum + right_sum + root.val
            no_of_nodes = nl + nr + 1

            avg = cur_sum / no_of_nodes

            # update ans
            ans = max(avg, ans)

            return (cur_sum, no_of_nodes)

        return ans

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