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Problem solution · C++

Maxpathsum

Maxpathsum: a C++ solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Depth-first search
Source
Kamyu LeetCode Solutions
Length
35 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Maxpathsum, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 35 lines of C++ from the credited upstream file maxPathSum.cpp.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaxpathsum · C++C++
Use this to learn the idea, then write your own version.
// Time Complexity: O(n)// Space Complexity: O(logn) /** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {    public:        int maxPathSum(TreeNode *root) {            max_sum = INT_MIN;            dfs(root);            return max_sum;         }    private:        int max_sum;        int dfs(const TreeNode *root) {            if(!root) return 0;            int l = dfs(root->left);            int r = dfs(root->right);            int sum = root->val;            if(l > 0) sum += l;            if(r > 0) sum += r;            max_sum = max(max_sum, sum);            return max(r, l) > 0? max(r, l) + root->val : root->val;        } };

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