<LeetCode><Easy> 112 Path Sum --二叉树深度优先遍历

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.

#Python2 76ms

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

class Solution(object):
def hasPathSum(self, root, sum):
"""
:type root: TreeNode
:type sum: int
:rtype: bool
"""
if not root:return False
def search(root,tmpSum=0):
if not root:
if tmpSum is sum:return True
else:return False
tmpSum+=root.val
if root.left:
if search(root.left,tmpSum):return True
if root.right:
if search(root.right,tmpSum):return True
if not root.left and not root.right and tmpSum==sum:return True
return False
return search(root)