Problem description:

Given the root of a binary tree, return the inorder traversal of its nodes’ values.
Example 1:


Input: root = [1,null,2,3]
Output: [1,3,2]

Solution:

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# 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: TreeNode) -> List[int]:
        if not root:
            return []
        stk = collections.deque()
        res = []
        
        while True:
            while root:
                stk.append(root)
                root = root.left
            if not stk:
                return res
            top = stk.pop()
            res.append(top.val)
            root = top.right
        return res
        
        
        '''
        if not root:
            return []
        
        def traverse(root, res):
            if not root:
                return None
            traverse(root.left, res)
            res.append(root.val)
            traverse(root.right, res)
        res = []
        traverse(root, res)
        return res
        '''

time complexity: $O(n)$
space complexity: $O(n)$
reference:
related problem: