【错题集-编程题】二叉树中的最大路径和（树形dp）

//值得学习的代码
class Solution
{
public:
    int ret = -1010;
 
    int maxPathSum(TreeNode* root) 
    {
        dfs(root);
        return ret;
    }
 
    int dfs(TreeNode* root)
    {
        if(root == nullptr) return 0;
 
        int l = max(0, dfs(root->left));// 左⼦树的最⼤单链和
        int r = max(0, dfs(root->right)); // 右⼦树的最⼤单链和
        // 经过root的最⼤路径和
        ret = max(ret, root->val + l + r);
 
        return root->val + max(l, r);
    }
};
 
//力扣AC代码
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
private:
    int maxSum=INT_MIN;
public:
    int maxGain(TreeNode* node)
    {
        if(node==nullptr) return 0;
        int leftGain=max(0, maxGain(node->left));
        int rightGain=max(0, maxGain(node->right));
        int sum=leftGain+rightGain+node->val;
        maxSum=max(maxSum, sum);
        return node->val+max(leftGain, rightGain);
    }
    int maxPathSum(TreeNode* root) {
        maxGain(root);
        return maxSum;
    }
};