Difficulty

Easy.

Problem

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.

AC

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



# DFS
class Solution():
    def hasPathSum(self, root, sum):
        if not root:
            return False
        if not root.left and not root.right:
            return root.val == sum
        else:
            return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.
 
val)