# Binary Tree Maximum Path Sum

#1

Given a non-empty binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root.

Example 1:

```Input: [1,2,3]

1
/ \
2   3

Output: 6
```

Example 2:

```Input: [-10,9,20,null,null,15,7]

-10
/ \
9  20
/  \
15   7

Output: 42
```

#2

We do a post order traversal and capture the max sum of all possible branches.

``````class Solution {
public int maxPathSum(TreeNode root) {
if(root==null) return 0;
int max[] = new int[1];
max[0] = Integer.MIN_VALUE;
helper(root, max);
return max[0];
}

private int helper(TreeNode root, int[] max){
if(root==null) return 0;
int leftMax = Math.max(0, helper(root.left, max)); //avoids negative values
int rightMax = Math.max(0, helper(root.right, max));
max[0] = Math.max(max[0], root.val+leftMax+rightMax); //considers the whole branch
// we already captured the current branch max, so we pick only left or right to make room for larger branches.
return root.val+Math.max(leftMax, rightMax);
}
}
``````