LeetCode: Binary Tree Maximum Path Sum

Title:

Given a binary tree, find the maximum path sum.

The path may start and end at any node in the tree.

For example:
Given the below binary tree,

       1
      / \
     2   3

 

Return 6.

 

思路：有点混乱，开始的想法是肯定使用递归，看左右子节点的值，和根值。另外，是求最大值，所以中间就存在最大值，比较左右和根值。但是代码写的很不好。学习好代码

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 == nullptr) 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;//值得学习
    }
};

 https://segmentfault.com/a/1190000003554858