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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.
1 # Definition for a binary tree node. 2 # class TreeNode(object): 3 # def __init__(self, x): 4 # self.val = x 5 # self.left = None 6 # self.right = None 7 8 class Solution(object): 9 def hasPathSum(self, root, sum): 10 """ 11 :type root: TreeNode 12 :type sum: int 13 :rtype: bool 14 """ 15 if not root: 16 return False 17 18 if not root.left and not root.right and root.val == sum: 19 return True 20 21 sum -= root.val 22 23 return self.hasPathSum(root.left, sum) or self.hasPathSum(root.right, sum)
112. Path Sum
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