<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:

    vector<vector<int>> res; //to store final result

    void traverse(TreeNode* root, vector <int> &path, int rem){
        if(!root) return; //return at NULL

        if(!root->left && !root->right){ //leaf node reached
            path.push_back(root->val); //do
            if(rem == root->val) res.push_back(path);
        //if remaining sum is same as value of root we have valid path
            path.pop_back(); // undo
            return;
        }

        int val = root->val; //subtract the contribution of this value from rem (remaining sum)
        path.push_back(val); //do
        traverse(root->left, path, rem-val); //recurse left subtree
        traverse(root->right, path, rem-val); //recurse right subtree
        path.pop_back(); //undo

    }

    vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
        res.clear(); //reset res 
        if(!root) return res; //if root itself NULL there are no paths
        vector <int> path; //to store the path
        traverse(root, path, targetSum);
        return res;
    }
};









 


  
        
    

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