leetcode 刷题 112 113

# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None

class Solution: def hasPathSum(self, root: TreeNode, sum: int) -> bool: # return False if not root else any(self.hasPathSum(n, sum-root.val) for n in [root.left, root.right]) if root.left or root.right else sum==root.val if not root: return False if not root.left and not root.right: return sum == root.val sum -= root.val return self.hasPathSum(root.left, sum) or self.hasPathSum(root.right, sum)

  

 

  
1
  
2
  
3
  
4
  
5
  
6
  
7
  
8
  
9
  
10
  
11
  
12
  
13
  
14
  
15
  
16
  
17
 

# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None

class Solution: def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]: ans = [] def dfs(r, tmp, s): if not r: return if not r.left and not r.right: if s == sum: ans.append(tmp) return if r.left: dfs(r.left, tmp + [r.left.val], s + r.left.val) if r.right: dfs(r.right, tmp + [r.right.val], s + r.right.val) if not root: return [] dfs(root, [root.val], root.val) return ans 

  

 

  
1
  
2
  
3
  
4
  
5
  
6
  
7
  
8
  
9
  
10
  
11
  
12
  
13
  
14
  
15
  
16
  
17
  
18
  
19
  
20
  
21
  
22
  
23
  
24
  
25
  
26
 

class Solution: def pathSum(self, root: TreeNode, sums: int) -> List[List[int]]: if not root: return [] if root.val==sums and not root.right and not root.left: return [[root.val]] right = self.pathSum(root.right,sums-root.val) left = self.pathSum(root.left, sums-root.val) res = [] while left: tmp = [root.val] tmp.extend(left.pop()) res.append(tmp) while right: tmp = [root.val] tmp.extend(right.pop()) res.append(tmp) return res

  

 

  
1
  
2
  
3
  
4
  
5
  
6
  
7
  
8
  
9
  
10
  
11
  
12
  
13
  
14
  
15
  
16
  
17
  
18
 

文章来源: maoli.blog.csdn.net，作者：刘润森！，版权归原作者所有，如需转载，请联系作者。

原文链接：maoli.blog.csdn.net/article/details/90723734