<p>Given the <code>root</code> of a binary tree and an integer <code>targetSum</code>, return <em>all <strong>root-to-leaf</strong> paths where the sum of the node values in the path equals </em><code>targetSum</code><em>. Each path should be returned as a list of the node <strong>values</strong>, not node references</em>.</p>

<p>A <strong>root-to-leaf</strong> path is a path starting from the root and ending at any leaf node. A <strong>leaf</strong> is a node with no children.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/pathsumii1.jpg" style="width: 500px; height: 356px;" />
<pre>
<strong>Input:</strong> root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
<strong>Output:</strong> [[5,4,11,2],[5,8,4,5]]
<strong>Explanation:</strong> There are two paths whose sum equals targetSum:
5 + 4 + 11 + 2 = 22
5 + 8 + 4 + 5 = 22
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/pathsum2.jpg" style="width: 212px; height: 181px;" />
<pre>
<strong>Input:</strong> root = [1,2,3], targetSum = 5
<strong>Output:</strong> []
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> root = [1,2], targetSum = 0
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 5000]</code>.</li>
	<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>
	<li><code>-1000 &lt;= targetSum &lt;= 1000</code></li>
</ul>

class Solution {
public:
    unordered_map<TreeNode*, bool> visited;
    vector<vector<int>> dfs(TreeNode* root, vector<int>& trace, int sum) {
        if (root == nullptr)
            return {};

        vector<vector<int>> ans;
        if (sum == root->val && root->left == nullptr && root->right == nullptr) {
            trace.emplace_back(sum);
            ans.emplace_back(trace);
            trace.pop_back();
            return ans;
        }

        int new_sum = sum - root->val;
        trace.emplace_back(root->val);
        auto left = dfs(root->left, trace, new_sum);
        auto right = dfs(root->right, trace, new_sum);
        trace.pop_back();
        ans.insert(ans.end(), left.begin(), left.end());
        ans.insert(ans.end(), right.begin(), right.end());
        return ans;
    }

    vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
        vector<int> trace;
        auto ans = dfs(root, trace, targetSum);
        if (ans.empty() || ans[0].empty())
            return {};
        return ans;
    }
};


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