114. Flatten Binary Tree to Linked List 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

前序遍历

将二叉树展开为单链表之后，单链表中的节点顺序即为二叉树的前序遍历访问各节点的顺序。因此，可以对二叉树进行前序遍历，获得各节点被访问到的顺序。由于将二叉树展开为链表之后会破坏二叉树的结构，因此在前序遍历结束之后更新每个节点的左右子节点的信息，将二叉树展开为单链表。

Code

 def flatten(self, root: TreeNode) -> None:"""Do not return anything, modify root in-place instead."""def preOrderTraversal(root: TreeNode):if root:preOrderList.append(root)preOrderTraversal(root.left)preOrderTraversal(root.right)preOrderList = []preOrderTraversal(root)size = len(preOrderList)for i in range(1, size):prev, curr = preOrderList[i - 1], preOrderList[i]prev.left = Noneprev.right = curr

复杂度分析

时间复杂度：O(n)，其中 n 是二叉树的节点数。前序遍历的时间复杂度是 O(n)，前序遍历之后，需要对每个节点更新左右子节点的信息，时间复杂度也是 O(n)。

空间复杂度：O(n)，其中 n 是二叉树的节点数。空间复杂度取决于栈（递归调用栈或者迭代中显性使用的栈）和存储前序遍历结果的列表的大小，栈内的元素个数不会超过 n，前序遍历列表中的元素个数是 nn。

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114. Flatten Binary Tree to Linked List 二叉树展开为链表

前序遍历

Code

复杂度分析

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