leetcode(44)-二叉树展开为链表

给定一个二叉树，原地将它展开为一个单链表。

例如，给定二叉树

    1
   / \
  2   5
 / \   \
3   4   6
将其展开为：

1
 \
  2
   \
    3
     \
      4
       \
        5
         \
          6

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/flatten-binary-tree-to-linked-list
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。```

我的思路，能往左走就往左走，走的时候把右子树存在栈里，走不动的时候，从栈里拿出来一个
```python
# 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 flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        stack = [None]
        tmp = root
        while True:
            if tmp is None:
                break
            if tmp.left is not None:
                if tmp.right is not None:stack.append(tmp.right)
                tmp.right = tmp.left
                tmp.left = None
                tmp = tmp.right
            elif tmp.left is None and tmp.right is not None:
                tmp = tmp.right
            elif tmp.left is None and tmp.right is None:
                head = stack.pop()
                tmp.right = head
                tmp = head

将右子树接在左子树的最左子节点上，然后转换

# 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 flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        if not root:
            return 
        while root:
            left, left_s, right = root.left, root.left, root.right
            if left:
                while left.right:
                    left = left.right
                left.right = right
                root.right = left_s
                root.left = None
            root = root.right