# Maximum sum along any tree path with positive and negative values

## Problem

You are given a binary tree with positive or negative numbers in its nodes. Find the maximum possible sum along any path of the binary tree.

## Solution

We do a postorder traversal, so move from leaf to node. Along any path we keep on adding the value to a local sum. If local sum is less than zero we discard that path and make local sum to zero and keep on continuing up. Every time we find a local sum we compare it with a global maximum and update the global whenever local is greater than global. At any node we check the maximum local sum of left and right children and take the larger of these two values in our local sum calculation. When we reach a null node we make our local sum zero and return from there.

## Code

public class MaxSumTreePathNegative { public static void main(String[] args) { Node a = new Node(4); Node b = new Node(-2); Node c = new Node(-1); Node d = new Node(6); Node e = new Node(-5); Node f = new Node(-2); Node g = new Node(7); Node h = new Node(8); a.left = b; a.right = c; b.left = d; c.left = e; c.right = f; f.left = g; f.right = h; int maxSum = findMaxSum(a); System.out.println(maxSum); } private static int findMaxSum(Node root) { int[] localMax = new int [1]; int[] globalMax = new int [1]; findMaxSum(root, localMax, globalMax); return globalMax[0]; } private static void findMaxSum(Node root, int[] localMax, int[] globalMax) { if (root == null) { localMax[0]=0; return; } localMax[0]=0; findMaxSum(root.left, localMax, globalMax); int leftLocal=localMax[0]; localMax[0]=0; findMaxSum(root.right, localMax, globalMax); int rightLocal=localMax[0]; int maxLocal=Math.max(leftLocal,rightLocal); if(maxLocal+root.value > 0) { localMax[0]=maxLocal+root.value; if(globalMax[0] < localMax [0]) globalMax[0] = localMax [0]; } else { localMax[0]=0; } } static class Node { Node left; Node right; int value; public Node(int value) { this.value = value; } } }