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BST to greater tree

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Problem statement

Given a binary search tree of integers with N number of nodes. Your task is to convert it to a Greater Tree.

A Greater Tree is formed when data of every node of the original BST is changed to the original node’s data plus the sum of all node’s data greater than or equal to it.

A binary search tree (BST) is a binary tree data structure which has the following properties.

• The left subtree of a node contains only nodes with data less than the node’s data.
• The right subtree of a node contains only nodes with data greater than the node’s data.
• Both the left and right subtrees must also be binary search trees.

For example:

Input:

Output:

Because only 5 is greater than 4 in the above BST, node 4 will be updated to 9 (4 + 5 = 9).
Nodes 3, 4, and 5 are greater than 2, hence node 2 will be modified to 14 (2 + 3 + 4 + 5 = 14). 
Node with data 5 will remain the same because there is no node in the BST with data greater than 5.
Nodes 2,3, 4, and 5 are greater than 1, hence node 1 will be modified to 15 (1 + 2 + 3 + 4 + 5 = 15).   
Nodes 4 and 5 are greater than 3, hence node 3 will be modified to 12 (3 + 4 + 5 = 12).

Detailed explanation ( Input/output format, Notes, Images )

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 only 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 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1

Output Format :

For each test case, print the level order traversal of the Greater Tree with node values separated by a single space. Print -1 for NULL nodes. 

Output for every test case will be denoted 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 <= 100
1 <= N <= 5000
0 <= data <= 10^6
where N is the number of nodes in the tree, T represents the number of test cases and data denotes data contained in the node of the binary tree.


Time Limit: 1 sec

Sample Input 1:

2
3 1 5 -1 2 -1 -1 -1 -1
9 6 10 4 7 -1 11 -1 -1 -1 -1 -1 -1

Sample Output 1:

 8 11 5 -1 10 -1 -1  -1 -1 
 30 43 21 47 37 -1 11 -1 -1 -1 -1 -1 -1

Explanation of the Sample Input1:

Here we have 2 test cases, hence there are 2 binary trees.

Test Case 1:

Node 5 is greater than node 3, hence node 3 modifies to 8 (5 + 3 = 8).

Nodes 2,3 and 5 are greater than node 1, hence node 1 modifies to 10 (1 + 5 + 3 + 2 = 11).

No node is greater than node 5, hence node 5 remains unchanged.

Nodes 3 and 5 are greater than node 2, hence node 2 modifies to 10 (5 + 3 + 2 = 10).




Test Case 2:

Nodes 10 and 11 are greater than node 9, hence node 9 modifies to 30 (10 + 11 + 9 = 30).

Nodes 7,9,11 and 10 are greater than node 6, hence node 6 modifies to 43 (7 + 9 + 11 + 10 + 6 = 43).

Node 11 is greater than node 10, hence node 10 modifies to 21 (10 + 11 = 21).

Nodes 6,7,9,11 and 10 are greater than node 4, hence node 4 modifies to 47 (7 + 9 + 11 + 10 + 6 + 4 = 47).

Nodes 9,10 and 11 are greater than node 7, hence node 7 modifies to 37 (10 + 11 + 9 + 7 = 37).

No node is greater than node 11, hence node 11 remains unchanged.

Hint 1

Hint 2

Hint 3

Think of naively updating one element/node at a time.

Approaches (3)

Approach 1

Approach 2

Approach 3

Brute Force

This is a basic brute force approach, in which for every single node, we will traverse the whole tree to find all the elements which are greater or equal to the current node.
First, we will store all the nodes in an array or a similar data structure.
Further, we will traverse the tree again and for each node of the tree, we will loop over the array and find the sum of all the elements which are greater than or equal to the current node.
We will modify the current node with the new value obtained, and hence the binary search tree gets converted to a greater tree.

Time Complexity

O(N^2), where N is the number of nodes in the binary tree.

For each node, we are traversing the whole tree with N nodes.

Space Complexity

O(N), where N is the number of nodes in the binary tree.

O(H) recursion stack space is used by the algorithm, where H is the height of BST. In the worst case (for skewed trees), H becomes N.

Also, we are storing the N nodes in an array/list. Thus, the overall space complexity is O(N).

(100% EXP penalty)

BST to greater tree

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