# Ninja's Encryption ll

Posted: 27 Mar, 2021

Difficulty: Moderate

#### Ninja is assigned with the duty of creating some problems on algebra or we can say questions on evaluating the expression. After creating those questions he has to send it so that the answer sheet according to the questions can be made. So he thinks someone might get his paper in between so he thinks of an encryption technique.

#### So in this encryption technique, he writes his expression in the form of a binary tree where leaf nodes represent the value and all other nodes represent the binary operators like ‘-’, ‘+’, ‘/’, ‘*’.

#### So now ninja has to find a way of decrypting his own technique as he has to tell the way of decrypting his own expression so that the answer team is able to solve his expression.

#### So your task is to write the code that can evaluate the given expression in the form of a binary tree where each node must have two children except leaf nodes and leaf nodes represent integer while other nodes represent binary operators like ‘-’, ‘+’, ‘/’, ‘*’.

##### Example:

```
Consider the following binary search tree so the evaluation of the expression is ( 4 * ( 5 - ( 2+ 7 ) ) ) = ‘-16’
```

#### Note:

```
You have been given the root of the tree.
```

##### Input Format:

```
The first line contains an integer ‘T' which denotes the number of test cases or queries to be run. Then the test cases follow.
The first line of each test case contains the elements of the tree in the level order form separated by a single space.
If any node does not have a left or right child, take -1 in its place. Refer to the example below.
Example:
Elements are in the level order form. The input consists of values of nodes separated by a single space in a single line. 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 :
```

```
1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1
Explanation :
Level 1 :
The root node of the tree is 1
Level 2 :
Left child of 1 = 2
Right child of 1 = 3
Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6
Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)
Level 5 :
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:
1 2 3 -1 5 6 7 -1 -1 -1 -1
```

##### Output Format :

```
For each test case, print the evaluation of each expression
Print output of each test case in a separate line.
```

##### Note:

```
You are not required to print anything explicitly. It has already been taken care of. Just implement the function.
```

##### Constraints:

```
1 <= T <= 10^2
1 <= N <= 3*10^3
0 <= data <= 10^3 or ( ‘+’, ‘-’, ‘*’, ‘/’ )
Where ‘T’ is the total number of test cases, ‘N’ is the number of nodes of the tree, and data is the value mentioned on the node.
Time Limit: 1 sec
```

Approach 1

The idea is that we can think of a postorder traversal of the tree as we know in postorder first we go to the left and right side of the node and in this case, only leaf nodes contain integers.

- So starting from the root node we start our postorder traversal i.e first we recursively call out the left tree and then the right tree.
- Now we make a base condition that whenever we encounter a node that has no children we simply convert it into its node value in the form of integers.
- Now after our two recursive calls as we know in postorder we arrive at the root node then we can check what binary operator it is storing.
- According to the binary operator in the node, we can return that required result.
- So in this way, we arrive at our final result.

#### Algorithm:

- If both node->left && node->right are NULL
- Then we can convert node->data into integer type
- Now we can recursively call our function postorder like
- postorder(node->left)
- postorder(node->right)

- Now we can evaluate our expression according to whatever operator our node is storing

SIMILAR PROBLEMS

# Preorder Traversal

Posted: 22 Jul, 2021

Difficulty: Easy

# Find All Subsets

Posted: 23 Jul, 2021

Difficulty: Easy

# Print All Subsets

Posted: 23 Jul, 2021

Difficulty: Easy

# Maximum Sum BST

Posted: 27 Jul, 2021

Difficulty: Hard

# Vertical Sum in BST

Posted: 27 Jul, 2021

Difficulty: Moderate