?做题思路or感想：

• 也是遍历一遍二叉树的路径，每次遍历一个节点就把targeSum减去对应的节点值，如果最后是叶子节点并且targeSum为0，则符合条件

  class Solution {
  public:
      bool flag = false;
      void dfs(TreeNode* cur, int sum, int targetSum) {
          //判断条件
          if (sum == targetSum && !cur->left && !cur->right) {
              flag = true;
              return;
          } else {
              //分别向左右子树遍历
              if (cur->left) dfs(cur->left, sum + cur->left->val, targetSum);
              if (cur->right) dfs(cur->right, sum + cur->right->val, targetSum);
          }
      }
      bool hasPathSum(TreeNode* root, int targetSum) {
          if (root == nullptr)return false;
          dfs(root, root->val, targetSum);
          if (flag)return true;
          else return false;
      }
  };