# path-sum-iii


Try it on <a href='https://leetcode.com/problems/path-sum-iii'>leetcode</a>

## Description
<div class="description">
<div><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>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>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>
</div>
</div>

## Solution(Python)
```Python
class Solution(object):
    def pathSum(self, root, sum):
        def dfs(sumHash, prefixSum, node):

            if not node:
                return 0

            # Sum of current path
            prefixSum += node.val

            # number of paths that ends at current node
            path = sumHash[prefixSum - sum]

            # add currentSum to prefixSum Hash
            sumHash[prefixSum] += 1

            # traverse left and right of tree
            path += dfs(sumHash, prefixSum, node.left) + dfs(
                sumHash, prefixSum, node.right
            )

            # remove currentSum from prefixSum Hash
            sumHash[prefixSum] -= 1

            return path

        # depth first search, initialize sumHash with prefix sum of 0, occurring once
        return dfs(collections.defaultdict(int, {0: 1}), 0, root)

```