将二叉树拆成链表

LeetCode题目地址

def flatten(self, root):
        # write your code here
        if root == None:
            return root
        #用根节点初始化栈
        # 然后出栈一个，就入栈右孩子，左孩子    出1入2
        stackList = [root]
        dummy = TreeNode(-1)
        cur = dummy
        while len(stackList) != 0:
            node = stackList.pop()
            cur.right = node
            cur = node
            if node.right:
                stackList.append(node.right)
            if node.left:
                stackList.append(node.left)
                node.left = None
                #不要忘记将左儿子标记为 null，否则你可能会得到空间溢出或是时间溢出
        return dummy.right