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# Time: O(n)

# Space: O(h), h is height of binary tree

#

# 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:

# @param root, a tree node

# @param sum, an integer

# @return a boolean

def hasPathSum(self, root, sum):

if root is None:

return False

if root.left is None and root.right is None and root.val == sum:

return True

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

if __name__ == "__main__":

root = TreeNode(5)

root.left = TreeNode(4)

root.right = TreeNode(8)

root.left.left = TreeNode(11)

root.left.left.right = TreeNode(2)

print Solution().hasPathSum(root, 22)