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.
1 # Definition for a binary tree node. 2 # class TreeNode: 3 # def __init__(self, x): 4 # self.val = x 5 # self.left = None 6 # self.right = None 7 8 class Solution: 9 def hasPathSum(self, root: TreeNode, sum: int) -> bool: 10 res = self.helper(root, sum) 11 return res 12 13 def helper(self, root, sum): 14 if root is None: 15 return False 16 if root.left is None and root.right is None: 17 if sum == root.val: 18 return True 19 else: 20 return False 21 left = self.helper(root.left, sum - root.val) 22 right = self.helper(root.right, sum - root.val) 23 return left or right