257. Binary Tree Paths

# Easy

Not easy. Hard to think. Two cases are needed to consider in helper function.

1. root doesn't have any child, then this path completes

2. root is empty, then this path is invalid.

List1 = [1, 2, 3], List2 = [2, 4, 6]. List1 + List2 = [1, 2, 3, 2, 4, 6]

So in the implementation below, List1 = ['1->2->5'], List2 = ['1->3'], then List1+List2 = [['1->2->5'], ['1->3']].

Python

class Solution:
    def binaryTreePaths(self, root: TreeNode) -> List[str]:
        # edge case
        if root == None:
            return []
        
        # regular case
        path = ""
        return self.helper(root, path)
    
    def helper(self, root, path):
        # stop condition
        if root == None:
            return []
        
        if root.left == None and root.right == None:
            return [path + str(root.val)]
        
        # regular case
        path = path + str(root.val) + "->"
        return self.helper(root.left, path) + self.helper(root.right, path)

Time complexity = $O(n)$ , traversal all nodes.

Second Tab