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[LeetCode] 112. Path Sum_Easy tag: DFS

2018-07-13 06:16  Johnson_强生仔仔  阅读(231)  评论(0)  编辑  收藏  举报

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \      \
7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

 

典型的DFS, 然后需要将目前的path sum一起append进入stack里面, 然后判断path sum 是否跟给的sum相等并且是leaf, 即可.

 

1. Constraints

1) can be empty

 

2. IDeas

DFS:     T:O(n)    S: T(n)

 

3. Code

3.1) iterable

 1 # iterable
 2 
 3 class Solution:
 4     def pathSum(self, root, s):
 5         if not root: return False
 6         stack =[(root, root.val)]
 7         while stack:
 8             node, ps = stack.pop()
 9             if ps == s and not node.left and not node.right:
10                 return True
11             if node.left:
12                 stack.append((node.left, ps + node.left.val))
13             if node.right:
14                 stack.append((node.right, ps + node.right.val))
15         return False

 

3.2) recursive way.

1 class Solution:
2     def pathSum(self, root, s):
3         if not root: return False
4         if s- root.val == 0 and not root.left and not root.right:
5             return True
6         return self.pathSum(root.left, s- root.val) or self.pathSum(root.right, s- root.val)

 

4. Test cases

1) edge case

2)   s= 9

      5
     / \
    4   8