<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 dfs(root):
            count = 0
            stack = deque()
            if not root:
                return 0
            stack.append([root, ['l','r'], [root.val]])
            if root.val == targetSum:
                count+=1
            while stack:
                # print([(i[0].val, i[1], i[2]) for i in stack])
                if not stack[-1][0]:
                    stack.pop()
                elif stack[-1][1][0] == 'l':
                    stack[-1][1] = ['r']
                    if stack[-1][0].left != None:
                        temp = stack[-1][0].left.val
                        temp_arr = [temp]
                        count+= temp==targetSum
                        for i in stack[-1][2]:
                            temp_arr.append(i + temp)
                            count+= temp_arr[-1]== targetSum
                        stack.append([stack[-1][0].left, ['l','r'], temp_arr])
                        
                elif stack[-1][1][0] == 'r':
                    if stack[-1][0].right != None:
                        temp = stack[-1][0].right.val
                        temp_arr = [temp]
                        count+= temp==targetSum
                        for i in stack[-1][2]:
                            temp_arr.append(i + temp)
                            count+= temp_arr[-1]== targetSum
                        stack.append([stack.pop()[0].right, ['l','r'], temp_arr])
                    else:
                        stack.pop()
                        
            return count
        return dfs(root)
        
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