python-leetcode-路径总和 III

437. 路径总和 III - 力扣（LeetCode）

# 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)