Flatten a Multilevel Doubly Linked List

You are given a doubly linked list which in addition to the next and previous pointers, it could have a child pointer, which may or may not point to a separate doubly linked list. These child lists may have one or more children of their own, and so on, to produce a multilevel data structure, as shown in the example below.

Flatten the list so that all the nodes appear in a single-level, doubly linked list. You are given the head of the first level of the list.


Input:  1---2---3---4---5---6--NULL          
Output: 1-2-3-7-8-11-12-9-10-4-5-6-NULL
# Definition for a Node.
class Node:
    def __init__(self, val, prev, next, child):
        self.val = val
        self.prev = prev
        self.next = next
        self.child = child

class Solution:
    def flatten(self, head: 'Node') -> 'Node':
        if not head:
            return None
        stack = list()
        cur = head
        while cur:
            if cur.child:
                if cur.next:
                    stack += [cur.next]
                cur.next = cur.child
                cur.next.prev = cur
                cur.child = None
            if not cur.next and stack:
                temp = stack.pop()
                cur.next = temp
                temp.prev = cur
            cur = cur.next
        return head

Flatten Binary Tree to Linked List

Given a binary tree, flatten it to a linked list in-place.

For example, given the following tree:

/ \
2 5
/ \ \
3 4 6

The flattened tree should look like:

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

class Solution:
    def flatten(self, root: TreeNode) -> None:
        Do not return anything, modify root in-place instead.
        node, stack = root, []
        while node:
            if node.right:
            node.right = node.left
            node.left = None
            if not node.right and stack:
                temp = stack.pop()
                node.right = temp
            node = node.right