# Maximum sum from a tree with adjacent levels not allowed

Given a binary tree with positive integer values. Find the maximum sum of nodes such that we cannot pick two levels for computing sum

```Examples:

Input : Tree
1
/ \
2   3
/
4
\
5
/
6

Output :11
Explanation: Total items we can take: {1, 4, 6}
or {2, 3, 5}. Max sum = 11.

Input : Tree
1
/   \
2     3
/      /  \
4       5     6
/  \     /     /
17  18   19    30
/     /  \
11    12   13
Output :89
Explanation: Total items we can take: {2, 3, 17, 18,
19, 30} or {1, 4, 5, 6, 11, 12, 13}.
Max sum from first set = 89.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Explanation: We know that we need to get item values from alternate tree levels. This means that if we pick from level 1, the next pick would be from level 3, then level 5 and so on. Similarly, if we start from level 2, next pick will be from level 4, then level 6 and so on. So, we actually need to recursively sum all the grandchildren of a particular element as those are guaranteed to be at the alternate level.

We know for any node of tree, there are 4 grandchildren of it.

```    grandchild1 = root.left.left;
grandchild2 = root.left.right;
grandchild3 = root.right.left;
grandchild4 = root.right.right;
```

We can recursively call the getSum() method in the below program to find the sum of these children and their grandchildren. At the end, we just need to return maximum sum obtained by starting at level 1 and starting at level 2.

```// Java code for max sum with adjacent levels
// not allowed
import java.util.*;

public class Main {

// Tree node class for Binary Tree
// representation
static class Node {
int data;
Node left, right;
Node(int item)
{
data = item;
left = right = null;
}
}

// Recursive function to find the maximum
// sum returned for a root node and its
// grandchildren
public static int getSumAlternate(Node root)
{
if (root == null)
return 0;

int sum = root.data;
if (root.left != null) {
sum += getSum(root.left.left);
sum += getSum(root.left.right);
}

if (root.right != null) {
sum += getSum(root.right.left);
sum += getSum(root.right.right);
}
return sum;
}

// Returns maximum sum with adjacent
// levels not allowed. This function
// mainly uses getSumAlternate()
public static int getSum(Node root)
{
if (root == null)
return 0;

// We compute sum of alternate levels
// starting first level and from second
// level.
// And return maximum of two values.
return Math.max(getSumAlternate(root),
(getSumAlternate(root.left) +
getSumAlternate(root.right)));
}

// Driver function
public static void main(String[] args)
{
Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.right.left = new Node(4);
root.right.left.right = new Node(5);
root.right.left.right.left = new Node(6);
System.out.println(getSum(root));
}
}
```

Output:

```11
```

Time Complexity : O(n)

Exercise: Try printing the same solution for a n-ary Tree rather than a binary tree. The trick lies in the representation of the tree.

This article is contributed by Ashish Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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