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]
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