python
# 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:
prefix_sum = defaultdict(int) # 记录前缀和出现的次数
prefix_sum[0] = 1 # 初始状态，前缀和为0出现一次
def dfs(node, curr_sum):
if not node:
return 0
curr_sum += node.val # 更新当前前缀和
count = prefix_sum[curr_sum - targetSum] # 计算符合条件的路径数
prefix_sum[curr_sum] += 1 # 记录当前前缀和
count += dfs(node.left, curr_sum) # 递归左子树
count += dfs(node.right, curr_sum) # 递归右子树
prefix_sum[curr_sum] -= 1 # 回溯，撤销当前节点的前缀和
return count
return dfs(root, 0)