leetcode94 Binary Tree Inorder Traversal

"""
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]
"""
"""
中序遍历题
解法一递归
"""
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution1:
    def inorderTraversal(self, root):
        if root == None:
            return []
        else:
            List = []
            self.order(root, List)  #bug,刚开始没有把List传进去
        return List
    def order(self, root, List):
        if root.left:
            self.order(root.left, List)
        List.append(root.val)
        if root.right:
            self.order(root.right, List)
"""
解法二：非递归
栈 思路：
（1）树先一直向左走到叶节点并将沿途的结点入栈 ；
（2）然后向右走一步，重复第一步操作。
注意要将沿途的节点进行入栈
"""
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
class Solution2(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root == None:
            return []
        List = []
        stack = []
        while root or stack:
            if root:
                stack.append(root)
                root = root.left
            else:
                root = stack.pop(-1)
                List.append(root.val)
                root = root.right
        return List
if __name__ == '__main__':
    S = Solution2()
    l1 = TreeNode(1)
    l2 = TreeNode(2)
    l3 = TreeNode(3)
    l4 = TreeNode(4)
    l5 = TreeNode(5)
    l6 = TreeNode(6)
    l7 = TreeNode(7)
    root = l1
    l1.left = l2
    l1.right = l3
    l2.left = l4
    l2.right = l5
    l3.left = l6
    l3.right = l7
    print(S.inorderTraversal(root))