LeetCode-in-All
437. Path Sum III
Medium
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
Solution
from typing import Optional
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
# 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:
def dfs(node: TreeNode, targetSum: int, curr_sum: int) -> None:
if not node:
return
curr_sum += node.val
self.count += self.prefix_sum.get(curr_sum - targetSum, 0)
self.prefix_sum[curr_sum] = self.prefix_sum.get(curr_sum, 0) + 1
dfs(node.left, targetSum, curr_sum)
dfs(node.right, targetSum, curr_sum)
self.prefix_sum[curr_sum] -= 1
self.count = 0
self.prefix_sum = {0: 1}
dfs(root, targetSum, 0)
return self.count