Leetcode刷题第九天-回溯

113：路径总和II

链接：113. 路径总和 II - 力扣（LeetCode）

root=[-2,null,-3],targetSum=-5莫要忘记负数情况😓

 1 # Definition for a binary tree node.
 2 # class TreeNode:
 3 #     def __init__(self, val=0, left=None, right=None):
 4 #         self.val = val
 5 #         self.left = left
 6 #         self.right = right
 7 class Solution:
 8     def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
 9         if(not root ):  return []
10         re=[]
11         self.backtracking(root,targetSum,0,re,[])
12         return re
13     def backtracking(self,cur,targetSum,sums,re,path):
14         if(not cur):    return
15         if(not cur.right and not cur.left):
16             sums+=cur.val
17             path.append(cur.val)
18             if(sums==targetSum):    re.append(path[:])
19             sums-=cur.val
20             path.pop()
21             return
22         sums+=cur.val
23         path.append(cur.val)
24         self.backtracking(cur.left,targetSum,sums,re,path)
25         self.backtracking(cur.right,targetSum,sums,re,path)
26         path.pop()
27         sums-=cur.val
pathSum

140:单词拆分II

链接：140. 单词拆分 II - 力扣（LeetCode）

💕还得是我妹妹，得高人指点，换个思路😍

 1 class Solution:
 2     def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
 3         if(not s):  return[]
 4         re=[]
 5         self.backtracking(s, wordDict,"",re)
 6         return re
 7     def backtracking(self,s,wordDict,path,re):
 8         if(not s):
 9             re.append(path[:len(path)-1])
10             return
11         for i in range(len(s)):
12             if(s[:i+1] in wordDict):
13                 self.backtracking(s[i+1:],wordDict,path+s[:i+1]+" ",re)
wordBreak

明天继续~~~~

posted @ 2024-02-07 21:09  小小小茹茹  阅读(15)  评论(0)    收藏  举报