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 # # 每次进入左子树和右子树,在叶子节点进行判断,只要有一个满足即可 class Solution: def hasPathSum(self , root: TreeNode, sum: int) -> bool: if not root: return False def dfs(r, target): target -= r.val if not r.left and not r.right: return target == 0 elif r.left and not r.right: return dfs(r.left, target) elif not r.left and r.right: return dfs(r.right, target) else: return dfs(r.left, target) or dfs(r.right, target) return dfs(root, sum)