Path Sum
题目
给予一个二叉树,和一个值su,寻找是否有一个从根节点到叶节点的和为su,有则返回True,没有为False。比如:
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
hasPathSum(self, root, 22)
将返回True
节点为:
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
解题思路
class Solution:
# @param {TreeNode} root
# @param {integer} sum
# @return {boolean}
def hasPathSum(self, root, su):
flag = False
if not root:
return flag
su -= root.val
if su == 0 and (not root.left) and (not root.right):
flag = True
return flag or self.hasPathSum(root.left, su) or self.hasPathSum(root.right, su)