# Binary Tree Maximum Path Sum

`Input: root = [1,2,3]Output: 6Explanation: The optimal path is 2 -> 1 -> 3 with a path sum of 2 + 1 + 3 = 6.`
`Input: root = [-10,9,20,null,null,15,7]Output: 42Explanation: The optimal path is 15 -> 20 -> 7 with a path sum of 15 + 20 + 7 = 42.`
• The number of nodes in the tree is in the range `[1, 3 * 104]`.
• `-1000 <= Node.val <= 1000`

## Recursion

First of all, let’s simplify the problem and implement a function `max_gain(node)` which takes a node as an argument and computes a maximum contribution that this node and one/zero of its subtrees could add.

• Initiate `max_sum` as the smallest possible integer and call `max_gain(node = root)`.
• Implement `max_gain(node)` with a check to continue the old path/to start a new path:
• Base case: if the node is null, the max gain is `0`.
• Call `max_gain` recursively for the node children to computing max gain from the left and right subtrees : `left_gain = max(max_gain(node.left), 0)` and
`right_gain = max(max_gain(node.right), 0)`.
• Now check to continue the old path or to start a new path. To start a new path would cost `price_newpath = node.val + left_gain + right_gain`. Update `max_sum` if it's better to start a new path.
• For the recursion return the max gain the node and one/zero of its subtrees could add to the current path : `node.val + max(left_gain, right_gain)`.
• Time complexity: O(N), where `N` is a number of nodes, since we visit each node not more than 2 times.
• Space complexity: O(H), where H is a tree height, to keep the recursion stack.

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