Given a binary tree, return all root-to-leaf paths.

Note: A leaf is a node with no children.

Example:

Input:

1
/   \
2     3
\
5

Output: ["1->2->5", "1->3"]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3


## Solutions

### 1. DFS-递归

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

class Solution:
def binaryTreePaths(self, root: TreeNode) -> List[str]:
if not root:
return []
res = []
self.dfs(root, '', res)
return res

def dfs(self, root, path, res):
if not root:
return
if path == '':
path = str(root.val)
else:
path += '->' + str(root.val)
if not root.left and not root.right:
res.append(path)
if root.left:
self.dfs(root.left, path, res)
if root.right:
self.dfs(root.right, path, res)
# Runtime: 36 ms, faster than 88.05% of Python3 online submissions for Binary Tree Paths.
# Memory Usage: 13.8 MB, less than 9.52% of Python3 online submissions for Binary Tree Paths.


### 2. DFS-迭代

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

class Solution:
def binaryTreePaths(self, root: TreeNode) -> List[str]:
if not root:
return []
res = []
stack = [('', root)]
while stack:
path, node = stack.pop()
if not node:
continue
if path == '':
path = str(node.val)
else:
path += '->' + str(node.val)
if not node.left and not node.right:
res.append(path)
if node.left:
stack.append((path, node.left))
if node.right:
stack.append((path, node.right))
return res
# Runtime: 36 ms, faster than 88.12% of Python3 online submissions for Binary Tree Paths.
# Memory Usage: 13.8 MB, less than 9.52% of Python3 online submissions for Binary Tree Paths.


### 3. BFS-迭代

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

class Solution:
def binaryTreePaths(self, root: TreeNode) -> List[str]:
if not root:
return []
res = []
queue = collections.deque()
queue.append(('', root))
while queue:
path, node = queue.popleft()
if not node:
continue
if path == '':
path = str(node.val)
else:
path += '->' + str(node.val)
if not node.left and not node.right:
res.append(path)
if node.left:
queue.append((path, node.left))
if node.right:
queue.append((path, node.right))
return res
# Runtime: 36 ms, faster than 88.12% of Python3 online submissions for Binary Tree Paths.
# Memory Usage: 13.9 MB, less than 9.52% of Python3 online submissions for Binary Tree Paths.