题解 | #二叉树中和为某一值的路径(一)#
二叉树中和为某一值的路径(一)
https://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 def dfs(root,sum): if not root: return False if root.val == sum and root.left is None and root.right is None: return root return dfs(root.left,sum-root.val) or dfs(root.right,sum-root.val) return dfs(root,sum)