113. Path Sum II
這一題比較困難的是，雖然是樹的遍歷加上回溯法。
# 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: TreeNode, targetSum: int) -> List[List[int]]:
result = []
def backtrack(node, targetSum, curr):
if not node:
return
curr.append(node.val)
if node.val == targetSum and not node.left and not node.right:
result.append(list(curr))
else:
backtrack(node.left, targetSum - node.val, curr)
backtrack(node.right, targetSum - node.val, curr)
curr.pop()
backtrack(root, targetSum, [])
return result