257. Binary Tree Paths

Easy LeetCode

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

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How to solve the problem

Recursion (DFS)

# 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)