tree July 20, 2022

Binary tree inorder traversal

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

We will traverse the tree with recursive DFS, as it is inorder traversal, we first traverse the left subtree then append the root, and lastly we traverse the 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 inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        res = []

        def traverse(root, res):
            if not root:
                return

            traverse(root.left, res)
            res.append(root.val)
            traverse(root.right, res)

        traverse(root, res)
        return res

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