給你二叉樹的根節點 root 和一個整數目標和 targetSum ，找出所有 從根節點到葉子節點 路徑總和等于給定目標和的路徑。葉子節點 是指沒有子節點的節點。示例 1： 輸入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22 輸出：[[5,4,11,2],[5,8,4,5]]示例 2： 輸入：root = [1,2,3], targetSum = 5 輸出：[]示例 3： 輸入：root = [1,2], targetSum = 0 輸出：[]# 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) -> List[List[int]]:# 方法一 遞歸def helper(root, sum, temp):if root == None:return Noneif root.left == None and root.right == None and root.val == sum:temp += [root.val]res.append(temp)helper(root.left, sum-root.val, temp + [root.val])helper(root.right, sum-root.val, temp + [root.val])res = []helper(root, targetSum, [])return res# 方法二if root == None:return []res = []temp = []stack = [(root, targetSum, temp)]while stack:node, sum, temp = stack.pop()if node.left == None and node.right == None and node.val == sum:temp += [node.val]res.append(temp)if node.left:stack.append((node.left, sum-node.val, temp + [node.val]))if node.right:stack.append((node.right, sum-node.val, temp + [node.val]))return res

總結

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