【LeetCode】113. Path Sum II

【Description】
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

  5
 / \
4   8

/ /
11 13 4
/ \ /
7 2 5 1
Return:

[
[5,4,11,2],
[5,8,4,5]
]

【Idea】
二叉树题基本是递归解万物？

easy版112，每次recursion少一个path node list的参数即可

【Solution】

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def __init__(self):
        self.result = []

    def pathSum(self, root: TreeNode, sum: int) -> bool:
        # DFS & recursion | 
        def recursion(root, path, cur_sum):
            if not root:
                return False 
            if not root.left and not root.right: 
                if cur_sum+root.val == sum:
                    path.append(root.val)
                    self.result.append(copy.deepcopy(path))  # 注意这里需要用深拷贝来append
                    # print(self.result)
                if path:
                    path.pop()  

            recursion(root.left, path+[root.val,], cur_sum+root.val)
            recursion(root.right, path+[root.val,], cur_sum+root.val)


        recursion(root, [], 0)
        return self.result