<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>

class Solution {
public:
    map<long long,int> cnt[1001];
    int n = 0; // n is the number of node
    void join(map<long long,int>& h1, map<long long,int>& h2){
        if (h1.size() < h2.size()) h1.swap(h2); // swap two set if one is smaller than other.
        for(auto h : h2){
            h1[h.first] += h.second;
            // else h1.insert(h);
        }
    }
    int dfs(TreeNode* node, long long d, int targetSum, int& ans){
        // d += node->val;
        cnt[n][d + node->val] = 1;
        int idx = n++;
        if (node->left != NULL){
            int lidx = dfs(node->left, d + node->val, targetSum, ans);
            join(cnt[idx], cnt[lidx]);
        }
        if (node->right != NULL){
            int ridx = dfs(node->right, d + node->val, targetSum, ans);
            join(cnt[idx], cnt[ridx]);
        }
        ans += cnt[idx][d + targetSum];
        return idx;
    }
    int pathSum(TreeNode* root, int targetSum) {
        if (root == NULL) return 0;
        int ans = 0;
        dfs(root, 0, targetSum, ans);
        return ans;
    }
};
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