113. 路径总和 II
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
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/path-sum-ii
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
代码:
from leetcode_python.utils import *
class Solution:
def __init__(self):
"""
测试case:
[]
0
[1,2]
1
"""
self.res = []
def dfs(self,root,route,remain):
if root:
route.append(root.val)
if remain==root.val and not root.left and not root.right:
self.res.append(route.copy())
else:
self.dfs(root.left,route.copy(),remain-root.val)
self.dfs(root.right,route.copy(),remain-root.val)
def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
self.dfs(root,[],targetSum)
return self.res
def hasPathSum_112_路径总和(self, root: Optional[TreeNode], targetSum: int) -> bool:
if not root:return False
targetSum -= root.val
if root.left is None and root.right is None:return targetSum==0
else: return self.hasPathSum_112_路径总和(root.left,targetSum) or self.hasPathSum_112_路径总和(root.right,targetSum)
def test(data_test):
s = Solution()
data = data_test # normal
# data = [list2node(data_test[0])] # list转node
return s.pathSum(*data)
def test_obj(data_test):
result = [None]
obj = Solution(*data_test[1][0])
for fun, data in zip(data_test[0][1::], data_test[1][1::]):
if data:
res = obj.__getattribute__(fun)(*data)
else:
res = obj.__getattribute__(fun)()
result.append(res)
return result
if __name__ == '__main__':
datas = [
[],
]
for data_test in datas:
t0 = time.time()
print('-' * 50)
print('input:', data_test)
print('output:', test(data_test))
print(f'use time:{time.time() - t0}s')
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备注:
GitHub:https://github.com/monijuan/leetcode_python
CSDN汇总:模拟卷Leetcode 题解汇总_卷子的博客-CSDN博客
可以加QQ群交流:1092754609
leetcode_python.utils详见汇总页说明
先刷的题,之后用脚本生成的blog,如果有错请留言,我看到了会修改的!谢谢!
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