题解 | #二叉树中和为某一值的路径(一)#

#     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:
        def dfs(root,res):
            if root is None:return False
            if root.val==res and root.left is None and root.right is None:
                return True
            return dfs(root.left,res-root.val) or dfs(root.right,res-root.val)
        result=dfs(root,sum)
        return result

    
    
    

这样是正确的

    def hasPathSum(self , root: TreeNode, sum: int) -> bool:
        def dfs(root,res):
            if root is None:return False
            if root.val==res and root.left is None and root.right is None:
                return True
            if root.left:
                return dfs(root.left,res-root.val) 
            if root.right:
                return dfs(root.right,res-root.val)
        result=dfs(root,sum)
        return result

这是错误的

大佬求解