112_PathSum.py

# 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.

# 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, sum):
		"""
		:type root: TreeNode
		:type sum: int
		:rtype: bool
		"""
		if not root:
			return False
		if not root.left and not root.right:
			if root.val == sum:
				return True

		return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)