思路：
后序遍历。
实现：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode() : val(0), left(nullptr), right(nullptr) {} 8 * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} 9 * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} 10 * }; 11 */ 12 13 class Solution 14 { 15 public: 16 int dfs(TreeNode* root, int& res) 17 { 18 int l = -1001, r = -1001; 19 if (root->left) l = dfs(root->left, res); 20 if (root->right) r = dfs(root->right, res); 21 res = max(res, max(l, r)); 22 int tmp = root->val; 23 if (l > 0) tmp += l; 24 if (r > 0) tmp += r; 25 res = max(res, tmp); 26 return max(max(l, r), 0) + root->val; 27 } 28 int maxPathSum(TreeNode* root) 29 { 30 if (root == NULL) return 0; 31 int res = -1001; 32 dfs(root, res); 33 return res; 34 } 35 };