Leetcode Path sum II

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

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

就是递归回溯法的应用。2到3星难度。

如下程序：

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();
	}

//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();
	}

Leetcode Path sum II

原文：http://blog.csdn.net/kenden23/article/details/19330875