# Amusement park

Posted: 25 Mar, 2021
Difficulty: Moderate

## PROBLEM STATEMENT

#### Input Format:

``````The first line of input contains an integer 'T' representing the number of test cases. Then the test cases follow.

The first line of each test case contains a fun number assigned to each ride in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place.

For example, the input for the tree depicted in the below image would be :
`````` ``````4
2 6
1 3 5 7
-1 -1 -1 -1 -1 -1
``````

#### Explanation:

``````Level 1 :
The root node of the tree is 4

Level 2 :
Left child of 4 = 2
Right child of 4 = 6

Level 3 :
Left child of 2 = 1
Right child of 2 = 3
Left child of 6 = 5
Right child of 6 = 7

Level 4 :
Left child of 1 = null (-1)
Right child of 1 = null (-1)
Left child of 3 = null (-1)
Right child of 3 = null (-1)
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 7 = null (-1)
Right child of 7 = null (-1)

The first not-null node(of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.

The input ends when all nodes at the last level are null(-1).
``````
##### Note:
``````The above format was just to provide clarity on how the input is formed for a given tree.

The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:

4 2 6 1 3 5 7 -1 -1 -1 -1 -1 -1
``````

#### Output Format:

``````For each test case, print the maximum fun that ninja can get from the amusement park.

The output for each test case is printed in a separate line.
``````

#### Constraints:

``````1 <= T <= 5
1 <= N <= 5* 10^3
1 <= DATA <= 10^4

Time limit: 1 sec
`````` Approach 1
• In this approach, we will check all cases one by one using recursion.
• If we do not take a particular ride then we have two choices:
• Either we can also leave its children to ride and move on.
• Or we can visit the children nodes and move on.
• But If we visit a particular ride then we cant ride it children rides.

So we use a recursive function that returns the maximum fun value that can be achievable.

By checking all conditions what if visited a node or not.

Algorithm:

``````Int maxFun(root, parentVisited )
{
If parentVisited = true that means parent node is already visited

then we have only one choice

return maxFun(root->left,false) + maxFun(root ->right, false)

If parentvisited = false that means parent node is not rided
Then we have two choice:

rided = root.data + maxFun(root.right, true) + maxFun(root.left, true)

notRided =  maxFun(root.right, false) + maxFun(root.left, false)

return max(rided, notRided)
}``````