二叉树中的最大路径和
/**
* 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) {
reverse(root);
return max_sum;
}
int reverse(TreeNode* root) {
if (root == nullptr) {
return 0;
}
int l_sum = max(0, reverse(root->left));
int r_sum = max(0, reverse(root->right));
//判断以当前节点为顶节点的最大路径和,并尝试更新全局变量
max_sum = max(max_sum, root->val+l_sum+r_sum);
//将当前节点作为中间节点往上传,此时要取消当前节点同时连接左右两棵树的情况
//要么传单独左,要么传单独右
return root->val + max(l_sum, r_sum);
}
};