**Largest Tree Sum Path - Google Top Interview Questions**

### Problem Statement :

Given a binary tree root, return the largest sum of any path between any two nodes. Constraints n ≤ 100,000 where n is the number of nodes in root Example 1 Input root = [-6, [5, null, null], [4, [7, [4, [2, null, null], null], [8, null, null]], [12, null, null]]] Output 31 Explanation The largest sum path is: [8, 7, 4, 12]

### Solution :

` ````
Solution in C++ :
int helper(Tree* root, int& ans) {
if (!root) return 0;
int a = helper(root->left, ans);
int b = helper(root->right, ans);
ans = max(ans, max(root->val + a + b, root->val));
return max(a + root->val, b + root->val);
}
int solve(Tree* root) {
int ans = INT_MIN;
helper(root, ans);
return ans;
}
```

` ````
Solution in Java :
import java.util.*;
/**
* public class Tree {
* int val;
* Tree left;
* Tree right;
* }
*/
class Solution {
private int ans = Integer.MIN_VALUE;
public int solve(Tree root) {
traversal(root);
return ans;
}
private int traversal(Tree root) {
if (root == null) {
return 0;
}
int left = traversal(root.left);
int right = traversal(root.right);
int onePath = Math.max(Math.max(left, right) + root.val, root.val);
int topSum = Math.max(onePath, left + right + root.val);
ans = Math.max(ans, topSum);
return onePath;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, root):
ans = -math.inf
def dfs(root):
if root is None:
return 0
left = dfs(root.left)
right = dfs(root.right)
single = max(left + root.val, right + root.val, root.val)
top = max(single, left + right + root.val)
nonlocal ans
ans = max(ans, top)
return single
dfs(root)
return ans
```

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