Compute tree right view and target-sum subarrays

You are given two separate programming tasks.

Task 1: Right-side view of a binary tree

You are given the root of a binary tree. Imagine standing on the right side of the tree and looking at it. You can see exactly one node at each depth level: the rightmost node at that level.

Write a function that returns a list (or array) of the values of the nodes that would be visible from the right side, ordered from top to bottom.

Example

Consider this binary tree:

       10
      /  \
     5    20
    / \     \
   3   7     30

From the right side, you see nodes with values: [10, 20, 30].

Input/Output

• Input: root node of a binary tree where each node has:
  • val : integer
  • left : left child pointer (or null)
  • right : right child pointer (or null)
• Output: An array (or list) of integers representing the visible node values from top to bottom.

Constraints

• Number of nodes n can be up to about 10^4 .
• Node values can be any 32-bit signed integers.
• The tree does not have to be balanced.

Task 2: Count subarrays with a given sum

You are given an integer array nums and an integer k.

Return the total number of contiguous subarrays whose elements sum exactly to k.

Example

• Input: nums = [1, 2, 3] , k = 3
  Valid subarrays are:
  • [1, 2] (sum = 3)
  • [3] (sum = 3) So the answer is 2 .

Input/Output

• Input:
  • An integer array nums of length n .
  • An integer k .
• Output: A single integer representing the number of contiguous subarrays whose sum equals k .

Constraints

• 1 <= n <= 10^5 (you should aim for an algorithm that is close to O(n)).
• Elements of nums can be negative, zero, or positive (32-bit signed integers).
• k is a 32-bit signed integer.

You may implement both functions in any language of your choice.