1
\$\begingroup\$

I'm posting my code for a LeetCode problem. If you'd like to review, please do so. Thank you for your time!

Problem

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

Inputs

[1,2,3]
[-10,9,20,null,null,15,7]
[-10,9,20,null,null,15,7,9,20,null,null,15,7]
[-10,9,20,null,null,15,7,9,20,null,null,15,720,null,null,15,7,9,20,null,null,15,7]
[-10,9,20,null,null,15,7,9,20,null,null,15,720,null,null,15,7,9,20,null,null,15,7999999,20,null,null,15,7,9,20,null,null,15,720,null,null,15,7,9,20,null,null,15,7]

Outputs

6
42
66
791
8001552

Code

#include <cstdint>
#include <algorithm>

struct Solution {
    int maxPathSum(TreeNode* root) {
        std::int_fast64_t sum = INT_FAST64_MIN;
        depthFirstSearch(root, sum);
        return sum;
    }

private:
    static std::int_fast64_t depthFirstSearch(
        const TreeNode* node,
        std::int_fast64_t& sum
    ) {

        if (!node) {
            return 0;
        }

        const std::int_fast64_t left = std::max(
                                           (std::int_fast64_t) 0,
                                           depthFirstSearch(node->left, sum)
                                       );
        const std::int_fast64_t right = std::max(
                                            (std::int_fast64_t) 0,
                                            depthFirstSearch(node->right, sum)
                                        );
        sum = std::max(sum, left + right + node->val);
        return std::max(left, right) + node->val;
    }
};

References

\$\endgroup\$

2 Answers 2

1
\$\begingroup\$

There's not much to say about your answer, it looks fine! One could quibble over the names of variables, maybe left and right could be named left_sum and right_sum for example, and you could've used auto for the type of those two variables. But other than that I think there is nothing that can be improved.

\$\endgroup\$
0
2
\$\begingroup\$

Not sure why you decided to use std::int_fast64_t over the common int that is used as the type of the tree nodes values.

But since you did, it would be more idiomatic to do at least:

static_cast<std::int_fast64_t>(0);

instead of

(std::int_fast64_t) 0;
\$\endgroup\$
2
  • 1
    \$\begingroup\$ Maybe it is better to avoid casting altogether. Casting implies you started with some different type. You can construct a zero of the proper type directly by writing: std::int_fast64_t(0) \$\endgroup\$
    – G. Sliepen
    Jul 23, 2020 at 10:00
  • \$\begingroup\$ Or create a named object static constexpr std::int_fast64_t fast_zero(0); \$\endgroup\$ Jul 23, 2020 at 18:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.