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]
]


## Solutions

### 1. DFS-递归

思路是一样的，

# Time Complexity: O(n)
# Sapce Complexity: O(1)
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
if not root:
return []
res = []
self.dfs(root, sum, [], res)
return res

def dfs(self, root, sum, path, res):
if not root.left and not root.right and sum == root.val:
path.append(root.val)
res.append(path)
cur_sum = sum - root.val
if root.left:
self.dfs(root.left, cur_sum, path + [root.val], res)
if root.right:
self.dfs(root.right, cur_sum, path + [root.val], res)
# Runtime: 44 ms, faster than 99.04% of Python3 online submissions for Path Sum II.
# Memory Usage: 19.1 MB, less than 6.90% of Python3 online submissions for Path Sum II.


### 2. 迭代

# Time Complexity: O(n)
# Sapce Complexity: O(1)
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
if not root:
return []
if not root.left and not root.right and sum == root.val:
return [[root.val]]
cur_sum = sum - root.val
tmp = self.pathSum(root.left, cur_sum) + self.pathSum(root.right, cur_sum)

return [[root.val] + path for path in tmp]
# Runtime: 56 ms, faster than 61.99% of Python3 online submissions for Path Sum II.
# Memory Usage: 15.6 MB, less than 37.93% of Python3 online submissions for Path Sum II.