[LeetCode] Path Sum II

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]
]

刚开始写的程序是这样的：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 class Solution {
public:
	vector<vector<int>> pathSum(TreeNode* root, int sum) {
		if (root == NULL)
			return ret;
		tmp.push_back(root->val);
		if ((root->left == NULL) && (root->right == NULL))
		{
			if (sum == accumulate(tmp.begin(), tmp.end(), 0))
				ret.push_back(tmp);
		}
		else
		{
			if (root->left!=NULL)
				pathSum(root->left, sum);
			if (root->right!=NULL)
				pathSum(root->right, sum);
		}
		tmp.pop_back();
		return ret;
	}
private:
	vector<vector<int>> ret;
	vector<int> tmp;
};

运行时间260ms 左右，看讨论区里的代码和我这类似的，只用16ms ...

最后发现,问题出现在返回值上！

看来copy constructor和析构的过程确实代价很大啊.....

改成下面的形式就快多了：

class Solution {
public:
	void pathSum_aux(TreeNode* root, int sum) {
		if (root == NULL)	return;
		tmp.push_back(root->val);
		if ((root->left == NULL) && (root->right == NULL))
		{
			if (sum == root->val)
				ret.push_back(tmp);
		}
		else
		{
			if (root->left != NULL)
				pathSum_aux(root->left, sum - root->val);
			if (root->right != NULL)
				pathSum_aux(root->right, sum - root->val);
		}
		tmp.pop_back();
	}

	vector<vector<int>> pathSum(TreeNode* root, int sum) {
		pathSum_aux(root, sum);
		return ret;
	}

 16ms左右...