Given the root of a binary tree and an integer targetSum, return the number of paths where the sum of the values along the path equals targetSum.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).

This problem can be solved using Depth-First Search (DFS) and Hashmap. Here is a Python solution:

class Solution:
    def pathSum(self, root, target):
        # define global return var
        self.numOfPaths = 0
        # 1st layer DFS to go through each node
        self.dfs(root, target)
        # return result
        return self.numOfPaths
    
    # define: traverse through the tree, at each treenode we will initiate a 2nd layer DFS to test if a path sum includes the answer
    def dfs(self, node, target):
        # exit condition
        if node is None:
            return 
        # 2nd layer DFS
        self.test(node, target) # we know that node is not None
        # next step
        self.dfs(node.left, target)
        self.dfs(node.right, target)
        
    # define: for a given node, DFS to find any path that sum == target, if find self.numOfPaths += 1
    def test(self, node, target):
        # exit condition
        if node is None:
            return
        if node.val == target:
            self.numOfPaths += 1
            
        # next step
        self.test(node.left, target-node.val)
        self.test(node.right, target-node.val)

This solution works by initiating a DFS from each node (1st layer DFS), then for each node we will test if there is a path (could start at any node that is the child of the current node and includes the current node) sum equals to the target (2nd layer DFS). If so, increment the global counter. The time complexity for this approach is O(N^2) for the worst case where the tree is a single path, and O(NlogN) for a balanced binary tree.