tree July 20, 2022

Binary tree preorder traversal

Time O(n) Space O(n) Open original problem

We will traverse the tree with recursive DFS, as it is preorder traversal, we first append the root, then the left subtree and then right subtree.

# 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 preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        res = []

        def traverse(root, res):
            if not root:
                return
            res.append(root.val)
            traverse(root.left, res)
            traverse(root.right, res)

        traverse(root, res)
        return res

Time Complexity: O(n)
Space Complexity: O(n)