Solution Explanation: Binary Tree Maximum Path Sum

This problem asks for the maximum sum of a path in a binary tree. A path can start and end at any node, and it doesn't need to pass through the root. The crucial understanding is that a maximum path may or may not include the current node as its root.

Approach:

The most efficient approach utilizes Depth-First Search (DFS) with a recursive helper function. This function explores the tree recursively, calculating the maximum path sum achievable at each node.

The recursive function, dfs(), performs the following:

1. Base Case: If the current node is null, it returns 0. This signifies the end of a branch.

2. Recursive Calls: It recursively calls dfs() on the left and right children.

3. Maximum Path through Current Node: It calculates the maximum path sum that includes the current node. This involves:

   • Getting the maximum sum from the left subtree (left). If left is negative, we set it to 0 because including a negative path would reduce the total sum.
   • Getting the maximum sum from the right subtree (right). Similarly, if it's negative, we set it to 0.
   • The maximum path through the current node is then root.val + left + right. This is updated in a global variable ans to keep track of the overall maximum path sum encountered so far.

4. Maximum Path from Current Node (to a parent): It returns the maximum path sum starting from the current node and extending upwards to a parent node. This is crucial because the maximum overall path may not use both left and right subtrees. The function returns root.val + max(left, right).

5. Main Function: The main function initializes a global variable ans to negative infinity and then calls dfs(root). Finally, it returns the value of ans.

Time and Space Complexity:

• Time Complexity: O(N) - The dfs() function visits each node in the tree exactly once. Therefore, the time complexity is linear with respect to the number of nodes (N).

• Space Complexity: O(H) - The space complexity is determined by the maximum depth of the recursion stack, which is proportional to the height (H) of the tree. In the worst case (a skewed tree), H can be equal to N. In an average case balanced tree, H is log(N).

Code Implementation (Python):

class Solution:
    def maxPathSum(self, root: Optional[TreeNode]) -> int:
        self.max_sum = float('-inf')  # Initialize with negative infinity
 
        def dfs(node):
            if not node:
                return 0
 
            left_sum = max(0, dfs(node.left))  # Ensure non-negative values
            right_sum = max(0, dfs(node.right))
 
            self.max_sum = max(self.max_sum, node.val + left_sum + right_sum)
            return node.val + max(left_sum, right_sum)  # Max path going up
 
        dfs(root)
        return self.max_sum

The other languages (Java, C++, Go, TypeScript, Rust, JavaScript, C#) follow a very similar structure; only syntax and standard library functions differ slightly. They all implement the same recursive DFS algorithm. Refer to the provided code snippets in the original response for the implementations in these languages.