<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

#fixed bug:Issues Identified
# Counting Paths Incorrectly: The logic for counting paths that sum to targetSum is incorrect. Specifically, the remain == 0 check should be counting the initial path from the root, but the remain in prefix_sum should be incrementing by the number of times remain appears in prefix_sum, not just by 1.
# Prefix Sum Handling: The prefix sums are being handled using a set, which is incorrect. We need to use a dictionary to keep track of the count of prefix sums.

#my solution w gpt debug(47%)
class Solution:
    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> int:
        count = 0
        def dfs(root, curr_sum, prefix_sum):
            nonlocal count
            if not root: return
            curr_sum += root.val
            remain = curr_sum - targetSum
            print(f'root.val{root.val}remain{remain}prefix{prefix_sum}')
            if remain == 0:
                count += 1
                print(count)
            if remain in prefix_sum:
                count += prefix_sum[remain]
                print(count)
            prefix_sum[curr_sum] += 1
            dfs(root.left, curr_sum, prefix_sum.copy())
            dfs(root.right, curr_sum, prefix_sum.copy())
        dfs(root, 0, defaultdict(int))
        return count

#best runtime
# class Solution:
#     def pathSum(self, root: Optional[TreeNode], targetSum: int) -> int:
#         self.res = 0
#         def traverse(node, curSum, d):
#             if not node:
#                 return
            
#             curSum += node.val

#             otherSum = targetSum - curSum
            
#             # print(otherSum, curSum, node, d)
#             if -otherSum in d:
#                 self.res += d[-otherSum]
#                 # print(self.res)
            
#             d[curSum] = d.get(curSum, 0) + 1
#             traverse(node.left, curSum, d)
#             traverse(node.right, curSum, d)
#             d[curSum] -= 1
#             curSum -= node.val
        
#         traverse(root, 0, {0: 1})
#         return self.res

            
               
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