题解 | #二叉树中和为某一值的路径(二)#

# 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 target int整型 
# @return int整型二维数组
#
class Solution:
    def findSubPath(self,root:TreeNode,path: List[int],paths:List[List[int]],target: int):       
        print(root.val,path) 
        path.append(root.val)
        if sum(path) == target:
            if root.left is None and root.right is None:
                paths.append(path)
        else:
            path_left = path
            path_right = path.copy()
            if root.left is not None :
                self.findSubPath(root.left,path_left,paths,target)
            if root.right is not None :
                self.findSubPath(root.right,path_right,paths,target)

    def FindPath(self , root: TreeNode, target: int) -> List[List[int]]:
        # write code here
        # 涉及深度遍历，需要递归
        paths=[]
        path = []
        if root is None:
            return []
        path.append(root.val)
        path_left = path
        path_right = path.copy()
        if sum(path) == target:
            if root.left is None and root.right is None:
                paths.append(path)
        if root.left is not None:
            self.findSubPath(root.left,path_left,paths,target)
        if root.right is not None:
            self.findSubPath(root.right,path_right,paths,target)
        return paths

需要注意的是，节点的值有可能是负数。