Maximum path sum from any node | # 87 GFG 160-Day Challenge| GFG POTD

Finding Maximum Path Sum from Any Node in a Binary Tree

The "Maximum Path Sum from Any Node" problem requires finding the highest sum possible in a binary tree, where the path can start and end at any node.

Optimal Approach: Recursion with Postorder Traversal

1. Use recursive traversal (postorder DFS) to compute the maximum sum from each node’s left and right subtrees.
2. Maintain two values for each node:

Single path sum: The maximum sum including the node and one of its subtrees (left or right).

Complete path sum: The sum when the node acts as a "bridge," including both left and right subtrees.
3. Update a global maximum to track the highest path sum found.
4. Return the single path sum to ensure valid subtree paths.
Example Walkthrough

Input:

10
/ \
2 10
/ \ \
20 1 -25
/ \
3 4

Processing:

The maximum path sum can be 20 → 2 → 10 → 10 = 42.

Another possible sum could be 3 → -25 → 4 = -18 (not optimal).

The result ignores paths that would decrease the sum.
Output:

Maximum Path Sum = 42

Use Cases in Real Life:

Circuit Network Optimization: Finding the most efficient route for power flow.

Game Development: Pathfinding for AI where rewards are maximized.

Finance: Finding the most profitable investment path in decision trees.
Useful Links:

Problem Link - https://www.geeksforgeeks.org/batch/gfg-160-problems/track/tree-gfg-160/problem/maximum-path-sum-from-any-node

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