题目

任意节点到另一任意节点的路径的权值最大是多少

思路

1. 如果当前节点的左右能连起来，则为node->left + node + node->right
2. 如果连不起来，返回给当前节点的父节点 node + max(left,right)

/**
 * 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 {
public:
    int max_sum = INT_MIN;
    int maxPathSum(TreeNode* root) {
        dfs(root);
        return max_sum;
    }
    int dfs(TreeNode* root){
        if(root == nullptr ) return 0 ;
        int left = max(0,dfs(root->left));
        int right = max(0,dfs(root->right));
        max_sum = max(max_sum,root->val + left + right);
        int ret = max(left,right) + root->val;
        return ret;
    }
};