题解 | #二叉树中和为某一值的路径(三)#

# 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 int整型
#

class Solution:
    def __init__(self) -> None:
        self.count = 0

    def dfs(self,root,sum):
        if root is None:
            return False
        else:
            if sum==root.val:
                self.count +=1
                
            leftf = self.dfs(root.left,sum-root.val)
            rightf = self.dfs(root.right,sum-root.val)
            return leftf or rightf

    def FindPath(self , root: TreeNode, sum: int) -> int:
        # write code here
        if root is None:
            return 0 
        else:
            get = self.dfs(root,sum)
            leftf = self.FindPath(root.left,sum)
            rightf = self.FindPath(root.right,sum)

            return self.count