Introduction

In this blog post, we will explore how to solve the LeetCode problem "437" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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). Example 1: Input: root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8 Output: 3 Explanation: The paths that sum to 8 are shown. Example 2: Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22 Output: 3 Constraints: The number of nodes in the tree is in the range [0, 1000]. -109 <= Node.val <= 109 -1000 <= targetSum <= 1000

Explanation

Here's the breakdown of the approach and the Python code:

• Prefix Sum Optimization: The core idea is to use a hash map to store the running prefix sums encountered during the traversal. This allows us to efficiently check if there exists a path ending at the current node that sums up to the targetSum. The difference between the current prefix sum and the targetSum represents the prefix sum we need to find in the path leading to the current node.

• DFS Traversal: We use Depth-First Search (DFS) to traverse the tree. As we move down the tree, we update the prefix sum. Before moving to a child, we check how many paths exist from the root to the current node that satisfy the condition (current sum - targetSum = previous sum). Then we explore the children and backtrack by removing the current sum from the hashmap after the exploration of the subtree.

• Hash Map for Path Count: The hash map stores the frequency of each prefix sum encountered along the current path. This enables us to quickly determine the number of paths with a specific sum.

• Time & Space Complexity: O(N), where N is the number of nodes in the tree. Space complexity is O(N) in the worst-case scenario (skewed tree) due to the recursion stack and the hash map.

Code

    class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def pathSum(root: TreeNode, targetSum: int) -> int:
    """
    Finds the number of paths in a binary tree that sum to a target value.

    Args:
        root: The root of the binary tree.
        targetSum: The target sum to find.

    Returns:
        The number of paths that sum to the target value.
    """

    count = 0
    prefix_sum_count = {0: 1}  # Initialize with a sum of 0 occurring once (for paths starting at the root)

    def dfs(node, current_sum):
        nonlocal count
        if not node:
            return

        current_sum += node.val

        # Check if there exists a path ending at the current node that sums to targetSum
        required_sum = current_sum - targetSum
        if required_sum in prefix_sum_count:
            count += prefix_sum_count[required_sum]

        # Update the count of the current sum
        prefix_sum_count[current_sum] = prefix_sum_count.get(current_sum, 0) + 1

        # Explore the left and right subtrees
        dfs(node.left, current_sum)
        dfs(node.right, current_sum)

        # Backtrack: Remove the current sum from the count as we move up the tree
        prefix_sum_count[current_sum] -= 1

    dfs(root, 0)
    return count