✅做题思路or感想：

• 也是遍历一遍二叉树，判断路径总和和是否是叶子节点，要注意的是这里还需要加一个vector来记录路径

• class Solution {
  public:
      vector<vector<int>>result;
      void dfs(TreeNode* cur, vector<int>vec, int sum) {
          //判断条件
          if (sum == 0 && !cur->left && !cur->right) {
              result.push_back(vec);
              return;
          } else {
              if (cur->left) {
                  //这里要有一个vector来记录路径
                  vec.push_back(cur->left->val);
                  dfs(cur->left, vec, sum - cur->left->val);
                  //回溯！
                  vec.pop_back();
              }
              if (cur->right) {
                  vec.push_back(cur->right->val);
                  dfs(cur->right, vec, sum - cur->right->val);
                  vec.pop_back();
              }
          }
      }
      vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
          if (root == nullptr)return result;
          dfs(root, vector<int>{root->val}, targetSum - root->val);
          return result;
      }
  };
  

标签：right,cur,val,back,II,vector,vec,113,总和
来源： https://www.cnblogs.com/doomaa/p/16069627.html