257. Binary Tree Paths
Easy LeetCodeGiven the root of a binary tree, return all root-to-leaf paths in any order.
A leaf is a node with no children.
Example 1
Input: root = [1,2,3,null,5]
Output: ["1->2->5","1->3"]
Example 2
Input: root = [1]
Output: ["1"]
Constraints
- The number of nodes in the tree is in the range [1, 100].
- -100 <= Node.val <= 100
How to solve the problem
Recursion (DFS)
python
# 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 binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
result = []
def dfs(node, path):
if node is None:
return
path += str(node.val)
if not node.left and not node.right:
result.append(path)
return
path += "->"
dfs(node.left, path)
dfs(node.right, path)
dfs(root, "")
return result
Complexity
- Time complexity: O(n), n == number of nodes
- Space complexity: O(n)
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