题解 | #二叉树中和为某一值的路径#

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    def __init__(self):
        self.paths = []
        self.path  = []
    def FindPath(self, root, expectNumber):
        # write code here
        if not root:
            return self.paths
        
        self.path.append(root.val)
#         print(self.path)

        if root.right is None and root.left is None:
            if sum(self.path) == expectNumber: 
                self.paths.append(self.path[::])
#             self.path.pop()
#             print("2:", self.path)
            return self.paths
        if root.left is not None:
            self.FindPath(root.left, expectNumber)
#             print(self.path)
            self.path.pop()
        if root.right is not None:
            self.FindPath(root.right, expectNumber)
            self.path.pop()
        return self.paths