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

 class Solution(object):
     def hasPathSum(self, root, sum):
         """
         :type root: TreeNode
         :type sum: int
         :rtype: bool
         """

         if not root:
             return False
         elif root.val == sum and not root.left and not root.right:
             return True
         else:
             return self.hasPathSum(root.right, sum - root.val) or self.hasPathSum(root.left, sum - root.val)

也可以使用DFS+backtracking

 class Solution(object):
     def hasPathSum(self, root, s):
         """
         :type root: TreeNode
         :type sum: int
         :rtype: bool
         """
         if not root:
             return False

         res = []
         self.dfs(root, s, [root.val], res)
         return any(res)

     def dfs(self, root, s, path, res):

         if not root.left and not root.right and sum(path) == s:
             res.append(True)

         if root.right:
             self.dfs(root.right, s, path+[root.right.val], res)
         if root.left:
             self.dfs(root.left, s, path+[root.left.val], res)