**Maximum Non-Adjacent Tree Sum - Amazon Top Interview Questions**

### Problem Statement :

Given a binary tree root, return the maximum sum of the integers that can be obtained given no two integers can be adjacent parent to child. Constraints n ≤ 100,000 where n is the number of nodes in root Example 1 Input root = [1, [4, [3, null, null], [2, null, null]], [5, null, null]] Output 10 Explanation We can pick 3, 2 and 5. Note if we picked 4, we wouldn't be able to pick 3 and 2 since they are adjacent.

### Solution :

` ````
Solution in C++ :
vector<int> helper(Tree* root, int& ans) {
if (root == NULL) {
return {0, 0};
}
if (root->left == root->right) {
ans = max(ans, root->val);
return {root->val, 0};
}
auto lef = helper(root->left, ans);
auto ri = helper(root->right, ans);
int taking = lef[1] + ri[1] + root->val;
int leaving = max(lef[0], lef[1]) + max(ri[0], ri[1]);
ans = max({ans, taking, leaving});
return {taking, leaving};
}
int solve(Tree* root) {
int ans = 0;
helper(root, ans);
return ans;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(Tree root) {
dfs(root);
return max;
}
int max = 0;
public int[] dfs(Tree root) {
if (root == null)
return new int[] {0, 0};
int[] left = dfs(root.left);
int[] right = dfs(root.right);
int[] res = new int[] {Math.max(left[0], left[1]) + Math.max(right[0], right[1]),
left[0] + right[0] + root.val};
max = Math.max(res[0], res[1]);
return res;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, root):
def dfs(node):
if node:
left_sum1, left_sum2 = dfs(node.left)
right_sum1, right_sum2 = dfs(node.right)
sum1 = left_sum2 + right_sum2 + node.val
sum2 = max(
left_sum1 + right_sum1,
left_sum1 + right_sum2,
left_sum2 + right_sum1,
left_sum2 + right_sum2,
)
return sum1, sum2
else:
return 0, 0
(sum1, sum2) = dfs(root)
return max(sum1, sum2)
```

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