Given a binary tree, return the inorder traversal of its nodes’ values.

Example:

Input: [1, null, 2, 3]

1
\
2
/
3


Output: [1, 3, 2]

Follow up: Recursive solution is trivial, could you do it iteratively?

## Solution

from typing import List

class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None

class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
list = []
stack = []

while root or stack:
while root:
stack.append(root)
root = root.left
root = stack.pop()
list.append(root.val)
root = root.right

return list


## Test Cases

test = Solution()
node1 = TreeNode(1)
node2 = TreeNode(2)
node3 = TreeNode(3)
node4 = TreeNode(4)
node5 = TreeNode(5)
node6 = TreeNode(6)
node7 = TreeNode(7)
node8 = TreeNode(8)
node9 = TreeNode(9)

answer = test.inorderTraversal(None)
assert answer == []
node1.right = node2
node2.left = node3
answer = test.inorderTraversal(node1)
assert answer == [1,3,2]

node1.left = node2
node1.right = node3
node2.left = node4
node2.right = node5
node5.left = node8
node3.left = node6
node3.right = node7
node6.right = node9
answer = test.inorderTraversal(node1)
assert answer == [4,2,8,5,1,6,9,3,7]
print('All Passed!')


## Big O Analysis

Space Complexity: O(N)

Time Complexity: O(N)