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.

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.

Something worth noticing in the problem:

a. It is a binary tree

b. only if there is a "root-to-leaf" path , is it a valid path

              5
             / \    given this tree and sum = 5, it is not a valid path since 
            4   8   it doesn't get down to the leaf.

It is clearly a recursive problem. Pass the current required sum to certain root until the leaf and see whether it is a valid path

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