<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>

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> int:

        def visit_fn2(node, cumsum, prefix_sums, cnt):
            if node is None: return cnt

            cumsum += node.val

            # the path all the way from root to node
            if cumsum == targetSum:
                cnt += 1

            try:
                cnt += prefix_sums[cumsum - targetSum]
            except KeyError:
                pass

            # increment the prefix sums
            prefix_sums[cumsum] = prefix_sums.get(cumsum, 0) + 1

            for c in [node.left, node.right]:
                cnt = visit_fn2(c, cumsum, { k: v for k, v in prefix_sums.items() }, cnt)

            return cnt

        def visit_fn(node, path_sums, cnt):
            if node is None:
                return cnt
            # extend all current paths to include node
            for i, s in enumerate(path_sums):
                path_sums[i] += node.val

            # add the path starting at node
            path_sums.append(node.val)

            # increment cnt by the number of paths down to node
            # which equal target
            cnt += sum(s == targetSum for s in path_sums)

            for c in [node.left, node.right]:
                cnt = visit_fn(c, [s for s in path_sums], cnt)
                
            return cnt
        
        # return visit_fn(root, [], 0)
        return visit_fn2(root, 0, {}, 0)
            
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