**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. Constraints 1 ≤ n ≤ 100,000 where n is the number of nodes in root Example 1 Input root = [1, [3, null, null], [7, [4, null, null], null]] Output 5.5 Explanation The subtree rooted at 7 has the highest average with (7 + 4) / 2. Example 2 Input root = [1, [2, null, null], [3, [4, null, null], null]] Output 4 Explanation 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)
dfs(root)
return ans
```

## View More Similar Problems

## Cycle Detection

A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer

View Solution →## Find Merge Point of Two Lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share

View Solution →## Inserting a Node Into a Sorted Doubly Linked List

Given a reference to the head of a doubly-linked list and an integer ,data , create a new DoublyLinkedListNode object having data value data and insert it at the proper location to maintain the sort. Example head refers to the list 1 <-> 2 <-> 4 - > NULL. data = 3 Return a reference to the new list: 1 <-> 2 <-> 4 - > NULL , Function Description Complete the sortedInsert function

View Solution →## Reverse a doubly linked list

This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.

View Solution →## Tree: Preorder Traversal

Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's

View Solution →## Tree: Postorder Traversal

Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the

View Solution →