[LC] 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.

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.

 # Definition for a binary tree node.
 # class TreeNode:
 #     def __init__(self, x):
 #         self.val = x
 #         self.left = None
 #         self.right = None

 class Solution:
     def hasPathSum(self, root: TreeNode, sum: int) -> bool:
         res = self.helper(root, sum)
         return res

     def helper(self, root, sum):
         if root is None:
             return False
         if root.left is None and root.right is None:
             if sum == root.val:
                 return True
             else:
                 return False
         left = self.helper(root.left, sum - root.val)
         right = self.helper(root.right, sum - root.val)
         return left or right