题解 |
二叉树中和为某一值的路径(一)
http://www.nowcoder.com/practice/508378c0823c423baa723ce448cbfd0c
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
#
# @param root TreeNode类
# @param sum int整型
# @return bool布尔型
#
class Solution:
def hasPathSum(self , root: TreeNode, sum: int) -> bool:
# write code here
if not root:
return False
if not (root.left or root.right):
if root.val==sum:
return True
else:
return False
else:
return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val)
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
#
# @param root TreeNode类
# @param sum int整型
# @return bool布尔型
#
class Solution:
def hasPathSum(self , root: TreeNode, sum: int) -> bool:
# write code here
if not root:
return False
if not (root.left or root.right):
if root.val==sum:
return True
else:
return False
else:
return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val)
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