Offer_day15_34. 二叉树中和为某一值的路径
输入一棵二叉树和一个整数,打印出二叉树中节点值的和为输入整数的所有路径。从树的根节点开始往下一直到叶节点所经过的节点形成一条路径。
示例:
给定如下二叉树,以及目标和 target = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
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返回:
[
[5,4,11,2],
[5,8,4,5]
]
提示:
节点总数 <= 10000
注意:本题与主站 113 题相同:https://leetcode-cn.com/problems/path-sum-ii/
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/er-cha-shu-zhong-he-wei-mou-yi-zhi-de-lu-jing-lcof
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:
import time
from typing import List
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution:
def __init__(self):
self.res = []
pass
def first(self, root:TreeNode, route:List[int], sum:int):
if root is None: return
sum+=root.val
route.append(root.val)
if sum==self.target and root.left is None and root.right is None:
self.res.append(route[:])
self.first(root.left,route,sum)
self.first(root.right,route,sum)
route.pop()
def pathSum(self, root: TreeNode, target: int) -> List[List[int]]:
self.target = target
self.first(root,[],0)
return self.res
def test(data_test):
s = Solution()
return s.getResult(*data_test)
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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