LeetCode (Python): Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?

Solution

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param root, a tree node
    # @return a list of integers
    def inorderTraversal(self, root):
        stack = []
        node = root
        solution = []
        while node!= None or len(stack)>0:
            if node != None:
                stack.append(node)
                node = node.left
            else:
                node = stack.pop()
                solution.append(node.val)
                node = node.right
        return solution

Other solution:

This solution needs more space but is easier to understand.

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param root, a tree node
    # @return a list of integers
    def inorderTraversal(self, root):
        stack = []
        if root != None:
            stack.append(root)
        solution = []
        used = set()
        while len(stack)>0:
            node = stack.pop()
            if node in used:
                solution.append(node.val)
            else:
                used.add(node)
                if node.right != None:
                    stack.append(node.right)
                stack.append(node)
                if node.left != None:
                    stack.append(node.left)
        return solution