**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
```

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