Leetcode 112. Path Sum

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.

For 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.

 

解法： 使用递归。在例子中，存在一个从root出发的，sum=22的路径 <=> 存在从4出发的，sum=22-4的路径 or 存在从8出发的，sum=22-8的路径

 1 class Solution(object):
 2     def hasPathSum(self, root, sum):
 3         """
 4         :type root: TreeNode
 5         :type sum: int
 6         :rtype: bool
 7         """
 8         
 9         if not root:
10             return False
11         elif root.val == sum and not root.left and not root.right:
12             return True
13         else:
14             return self.hasPathSum(root.right, sum - root.val) or self.hasPathSum(root.left, sum - root.val)

也可以使用DFS+backtracking

 

 1 class Solution(object):
 2     def hasPathSum(self, root, s):
 3         """
 4         :type root: TreeNode
 5         :type sum: int
 6         :rtype: bool
 7         """
 8         if not root:
 9             return False
10         
11         res = []
12         self.dfs(root, s, [root.val], res) 
13         return any(res)
14     
15     def dfs(self, root, s, path, res):
16 
17         if not root.left and not root.right and sum(path) == s:
18             res.append(True)
19         
20         if root.right:
21             self.dfs(root.right, s, path+[root.right.val], res)
22         if root.left:
23             self.dfs(root.left, s, path+[root.left.val], res)
24             
25

 

posted @ 2016-12-27 10:54  lettuan  阅读(134)  评论(0)    收藏  举报