Leetcode Path sum II

Path Sum II  
Total Accepted: 6531 
Total Submissions: 24225 My Submissions
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example:
Given the below binary tree and  sum = 22 ,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

再帰遡及法の応用である.2から3つ星の難易度.
次の手順に従います.

vector<vector<int> > pathSum(TreeNode *root, int sum) 
	{
		vector<vector<int> > rs;
		vector<int> tmp;
		storeSums(rs, tmp, root, sum);
		return rs;
	}
	void storeSums(vector<vector<int> > &rs, vector<int> &tmp, TreeNode *r, int sum)
	{
		if (!r) return;
		tmp.push_back(r->val);
		storeSums(rs, tmp, r->left, sum - r->val);
		storeSums(rs, tmp, r->right, sum - r->val);
		if (!r->left && !r->right && r->val == sum) rs.push_back(tmp);
		tmp.pop_back();
	}

次のような書き方は、多くの再帰遡及プログラムの標準形式と言える.

//2014-2-17 update
	vector<vector<int> > pathSum(TreeNode *root, int sum) 
	{
		vector<vector<int> > rs;
		vector<int> tmp;
		path(rs, tmp, root, sum);
		return rs;
	}
	void path(vector<vector<int> > &rs, vector<int> &tmp, TreeNode *r, int sum)
	{
		if (!r) return;
		if (!r->left && !r->right)
		{
			if (r->val == sum)
			{
				rs.push_back(tmp);
				rs.back().push_back(sum);
			}
			return;
		}
		tmp.push_back(r->val);
		path(rs, tmp, r->left, sum - r->val);
		path(rs, tmp, r->right, sum - r->val);
		tmp.pop_back();
	}