113. 路径总和 II

芒果香菠菠

于 2024-01-17 15:01:43 发布

阅读量364

点赞数 9

文章标签： python

本文链接：https://blog.youkuaiyun.com/weixin_42396719/article/details/135650898

版权

本文介绍了一个名为Solution的类，其中包含方法`pathSum`，用于在给定二叉树中找到所有节点值之和等于目标值`targetSum`的路径。方法利用`traversal`递归遍历树结构，记录路径并返回满足条件的路径列表。

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
    #     if not root:
    #         return []
    #     path = []
    #     return self.traversal(root, targetSum - root.val, path)
    # def traversal(self, root, count, path):
    #     if not root.left and not root.right and count == 0:
    #         return 1
    #     if not root.left and not root.right:
    #         return 0
    #     if root.left:
    #         count -= root.left.val
    #         if self.traversal(root.left, count, )     

        result = []
        self.traversal(root, targetSum, [], result)
        return result
    def traversal(self, node, count, path, result):
        if not node:
            return
        path.append(node.val)
        count -= node.val
        if not node.left and not node.right and count == 0:
            result.append(list(path))
        self.traversal(node.left, count, path, result)
        self.traversal(node.right, count, path, result)
        path.pop()