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
- 如果当前节点的左右能连起来,则为node->left + node + node->right
- 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;
}
};