题干：

代码;

class Solution {
public:bool traversal(TreeNode* node, int count){if(node == NULL)return false;if(!node -> left && !node -> right && count == 0)return true;if(!node -> left && !node -> right && count != 0)return false;if(node -> left){count -= node -> left -> val;if(traversal(node -> left, count))return true;count += node -> left -> val;}if(node -> right){count -= node -> right -> val;if(traversal(node -> right, count))return true;count += node -> right -> val;}return false;}bool hasPathSum(TreeNode* root, int targetSum) {if(root == NULL)return false;return traversal(root, targetSum - root -> val);}
};

注意count是从父节点开始减的，因为需要不断递归下去所以必须显示出方向，也即node -> left -> val。

另外必须判断root == NULL 情况，不管是在递归函数中还是在主函数中。

类似题：

代码：

class Solution {
private:void traversal(TreeNode* node, int count, vector<int> tmp, vector<vector<int>> &res){if(node == NULL)return;if(!node -> left && !node -> right && count == 0)res.push_back(tmp);if(!node -> left && !node -> right && count != 0)return;if(node -> left){tmp.push_back(node -> left -> val);count -= node -> left -> val;traversal(node -> left, count, tmp, res);count += node -> left -> val;tmp.pop_back();}if(node -> right){tmp.push_back(node -> right -> val);count -= node -> right -> val;traversal(node -> right, count, tmp, res);count += node -> right -> val;tmp.pop_back();}return;}
public:vector<vector<int>> pathSum(TreeNode* root, int targetSum) {vector<int> tmp;vector<vector<int>> res;if(root == NULL)return res;tmp.push_back(root -> val);traversal(root, targetSum - root -> val, tmp, res);return res;}
};

注意，需要在调用前先将根节点值传入tmp内，以及在调用时传入sum - root - > val。