Binary Tree Paths

Given 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

SOLUTION:

# 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]:
        op = []
        paths = [[root]]
        while len(paths) > 0:
            curr = paths.pop()
            if curr[-1].left or curr[-1].right:
                if curr[-1].left:
                    paths.append(curr + [curr[-1].left])
                if curr[-1].right:
                    paths.append(curr + [curr[-1].right])
            else:
                op.append(curr)
        return ["->".join([str(node.val) for node in path]) for path in op]