[数据结构与算法][LeetCode]113. 路径总和 II

113. 路径总和 II

难度：中等

给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。

叶子节点 是指没有子节点的节点。

示例 1：

输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出：[[5,4,11,2],[5,8,4,5]]

示例 2：

输入：root = [1,2,3], targetSum = 5
输出：[]

示例 3：

输入：root = [1,2], targetSum = 0
输出：[]

提示：

• 树中节点总数在范围 [0, 5000] 内
• -1000 <= Node.val <= 1000
• -1000 <= targetSum <= 1000

自解

# 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]]:
        if not root:
            return []

        def func(node,temp_result,value):
            if not node.left and not node.right:
                if value == targetSum:
                    result.append(temp_result)
                return
            if node.left:
                func(node.left,temp_result + [node.left.val],value  + node.left.val)
            if node.right:
                func(node.right,temp_result + [node.right.val],value  + node.right.val)

        temp_result = []
        result = []

        func(root,temp_result + [root.val],root.val)
        return result

官解

# 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 = []
        temp_result = []
        def func(root,value):
            if not root:
                return
            temp_result.append(root.val)
            value += root.val
            if not root.left and not root.right and value == targetSum:
                result.append(temp_result[:])
            func(root.left,value)
            func(root.right,value)
            temp_result.pop()
        func(root,0)
        return result