Find a highest sumGiven a binary tree in which each node element contains a number. Find the maximum possible path sum from one special node to another special node.Note: Here special node is a node that is connected to exactly one different node.Constraints:2  ≤  Number of nodes  ≤  104-103  ≤ Value of each node ≤ 103Sample input3 4 5 -10 4Sample output16

The problem is asking to find the maximum possible path sum from one special node to another special node in a binary tree. A special node is defined as a node that is connected to exactly one different node.

Here are the steps to solve this problem:

1. First, we need to understand the structure of the binary tree. In a binary tree, each node has at most two children, referred to as the left child and the right child.

2. We need to traverse the binary tree. We can do this using Depth-First Search (DFS) or Breadth-First Search (BFS). DFS is usually preferred for tree traversal.

3. While traversing, we need to keep track of the maximum path sum. We can do this by maintaining a variable, max_sum. We initialize it to the smallest possible value.

4. For each node, we calculate the maximum path sum that includes this node and update max_sum if it's greater than the current max_sum.

5. The maximum path sum that includes a node is calculated as the sum of the node's value and the maximum path sum of its left and right child. However, if the maximum path sum of a child is negative, we ignore it because it would decrease the total sum.

6. After traversing all nodes, max_sum will hold the maximum possible path sum from one special node to another special node.

7. Return max_sum as the result.

In Python, the code would look something like this:

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

def find_max_sum(root):
    max_sum = float('-inf')

    def helper(node):
        nonlocal max_sum
        if not node:
            return 0
        left = max(helper(node.left), 0)
        right = max(helper(node.right), 0)
        max_sum = max(max_sum, node.val + left + right)
        return node.val + max(left, right)

    helper(root)
    return max_sum

In the sample input, the binary tree is [3, 4, 5, -10, 4]. The maximum possible path sum from one special node to another special node is 16, which is the sum of 3, 4, and 5, and 4.