path-sum.py

# 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)