<p>Given the <code>root</code> of a binary tree and an integer <code>targetSum</code>, return <em>the number of paths where the sum of the values&nbsp;along the path equals</em>&nbsp;<code>targetSum</code>.</p>

<p>The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/09/pathsum3-1-tree.jpg" style="width: 450px; height: 386px;" />
<pre>
<strong>Input:</strong> root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
<strong>Output:</strong> 3
<strong>Explanation:</strong> The paths that sum to 8 are shown.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
<strong>Output:</strong> 3
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 1000]</code>.</li>
	<li><code>-10<sup>9</sup> &lt;= Node.val &lt;= 10<sup>9</sup></code></li>
	<li><code>-1000 &lt;= targetSum &lt;= 1000</code></li>
</ul>

#define ll long long

class Solution {

private:
vector<ll> childsum(TreeNode* node, ll target, ll &ans){
    
    if(!node) return {0};

    vector<ll> left, right, res;
    if(node->left) left = childsum(node->left, target, ans);
    if(node->right) right = childsum(node->right, target, ans);

    res.push_back(node->val);
    if(node->val == target) ans++;
    for(auto it : left) {
        res.push_back(it+node->val);
        if(it+ (ll)node->val == target) ans++;
    }
    for(auto it : right) {
        res.push_back(it + node->val);
        if(it+ (ll)node->val == target) ans++;
    }

    return res;
}
public:
    int pathSum(TreeNode* root, int target) {
        ll ans = 0;
        vector<ll> v = childsum(root, target, ans);
        return ans;
    }
};
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