Path Sum Iii

To solve this coding challenge, you need to implement a function that returns the number of paths in a binary tree where the sum of the node values in the path equals a given target sum. Below is a detailed explanation along with pseudocode to guide you through the solution.

Explanation:

We will approach this problem by using Depth-First Search (DFS). Here's a breakdown of the methodology:

1. Recursive DFS Traversal :
• We will perform a DFS traversal starting from each node.
• As we traverse, we calculate the current path sum.
• If the current path sum equals the target sum, we count this path as valid.
2. Subtree Traversal :
• For each node, not only do we consider paths starting from that node, but we also need to start new paths from both its left and right children.
• This ensures that we consider all possible paths in the tree.
3. Base Case :
• If the current node is
  None
  , we return 0 since there are no paths in an empty subtree.
4. Accumulating Path Counts :
• We need to keep a count of paths that sum to the target.
• Sum the paths found from the current node and the paths from the left and right subtrees.

Here's the detailed pseudocode with comments for clarity:

                                            
# Helper function for DFS traversal
FUNCTION dfs(currentNode, remainingSum):
    # Base case: if current node is None, return 0 as there are no paths
    IF currentNode IS None:
        RETURN 0
    
    # Initialize path count
    pathCount = 0
    
    # Check if the current node value matches the remaining sum
    IF currentNode.value == remainingSum:
        pathCount = pathCount + 1  # Found a valid path
    
    # Recursively traverse the left subtree with the updated remaining sum
    pathCount = pathCount + dfs(currentNode.left, remainingSum - currentNode.value)
    
    # Recursively traverse the right subtree with the updated remaining sum
    pathCount = pathCount + dfs(currentNode.right, remainingSum - currentNode.value)
    
    # Return the total count of valid paths from this node
    RETURN pathCount


# Main function to calculate the number of valid paths
FUNCTION pathSum(root, targetSum):
    # Base case: if the root is None, return 0 as there are no paths
    IF root IS None:
        RETURN 0
    
    # Calculate the number of valid paths starting from the root
    rootPathCount = dfs(root, targetSum)
    
    # Recursively calculate the number of valid paths for the left subtree
    leftPathCount = pathSum(root.left, targetSum)
    
    # Recursively calculate the number of valid paths for the right subtree
    rightPathCount = pathSum(root.right, targetSum)
    
    # Return the total number of valid paths
    RETURN rootPathCount + leftPathCount + rightPathCount

                                        

Step-by-Step Explanation:

• First, the
  dfs
  function is defined to count all valid paths starting from the given
  currentNode
  . It takes the
  currentNode
  and the
  remainingSum
  as arguments.
• Inside the
  dfs
  function:
• If the
  currentNode
  is
  None
  , it returns 0.
• Otherwise, it initializes a counter
  pathCount
  to zero and checks if the
  currentNode
  's value equals the
  remainingSum
  . If so, it increments
  pathCount
  .
• It then recursively calls itself for the left and right children of the
  currentNode
  , subtracting the
  currentNode
  's value from the
  remainingSum
  .
• Finally, it returns the total count of valid paths found.
• The
  pathSum
  function:
• Checks if the
  root
  is
  None
  . If so, it returns 0.
• Otherwise, it initializes the counts of valid paths starting from the
  root
  , left subtree, and right subtree by calling the
  dfs
  and
  pathSum
  functions recursively.
• It returns the sum of these counts.

This approach ensures that all possible paths in the binary tree are considered, and it counts every valid path where the node values sum up to the

targetSum

. The recursive nature helps traverse the tree in-depth for each path starting from every node.