This problem looks busy at first, but the accepted solution is built around one steady invariant. For 124. Binary Tree Maximum Path Sum, the solution in this repository is mainly a two pointers solution.
Guide
What?
Before optimizing anything, pin down what information is still useful after each move. Instead of trying to be clever immediately, read the code as a sequence of questions:
- What state are we keeping?
- How do we move from one state to the next?
- When do we know the answer is already determined?
For this file, the main tools are: two pointers.
The notes already sitting in the source point us in the right direction:
- https://leetcode.com/problems/binary-tree-maximum-path-sum/discuss/39775/Accepted-short-solution-in-Java
Guide
When?
Reach for this shape when a direct simulation would work logically but waste time revisiting the same information. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are maxPathDown, maxPathSum.
Guide
Why?
The code is doing bookkeeping so your brain does not have to keep the entire search space open at once.
- The final return is not magic; it is the invariant after the loops or recursion have finished doing their accounting.
Guide
How?
Walk through the solution in this order:
- Start from the smallest reliable state.
- Expand one legal move at a time.
- Cache, count, or merge information as soon as it becomes settled.
- Let the final stored value answer the original question.
The most important competitive-programming habit here is to trust the invariant. Once the invariant is right, the loops become much less scary.
Guide
Complexity
- Time: O(n) to O(n log n), depending on the dominant loop or data structure operation
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
01//https://leetcode.com/problems/binary-tree-maximum-path-sum/discuss/39775/Accepted-short-solution-in-Java
02//Runtime: 36 ms, faster than 33.76% of C++ online submissions for Binary Tree Maximum Path Sum.
03//Memory Usage: 28.8 MB, less than 6.06% of C++ online submissions for Binary Tree Maximum Path Sum.
04/**
05 * Definition for a binary tree node.
06 * struct TreeNode {
07 * int val;
08 * TreeNode *left;
09 * TreeNode *right;
10 * TreeNode() : val(0), left(nullptr), right(nullptr) {}
11 * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
12 * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
13 * };
14 */
15class Solution {
16public:
17 int ans;
18
19 int maxPathDown(TreeNode* node){
20 if(!node) return 0;
21 //max sum of path rooted at node->left, going down
22 int left = max(maxPathDown(node->left), 0);
23 //max sum of path rooted at node->right, going down
24 int right = max(maxPathDown(node->right), 0);
25 ans = max(ans, left+right+node->val);
26 //max sum of path that goes down
27 return node->val + max(left, right);
28 };
29
30 int maxPathSum(TreeNode* root) {
31 ans = INT_MIN;
32
33 maxPathDown(root);
34
35 return ans;
36 }
37};Cost