Leetcode Path sum II

最新推荐文章于 2017-08-12 17:16:16 发布

最新推荐文章于 2017-08-12 17:16:16 发布 · 228 阅读

本文详细介绍了使用递归回溯法解决二叉树中找到所有从根节点到叶子节点的路径，使得路径上的元素之和等于给定目标值的问题。通过给出具体的例子和代码实现，深入浅出地解释了该算法的核心思想和应用。

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]
]

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();
	}

下面这样写法可以说是很多递归回溯程序的标准形式了：

//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();
	}