feat: add solutions to lc problem: No.0430. Flatten a Multilevel Doubly Linked List

Author: yanglbme

solution/0100-0199/0160.Intersection of Two Linked Lists/README.md

@@ -18,7 +18,7 @@
 
 <p><strong>示例 1：</strong></p>
 
-<p><a href="https://assets.leetcode.com/uploads/2018/12/13/160_example_1.png" target="_blank"><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0160.Intersection%20of%20Two%20Linked%20Lists/images/160_example_1.png" style="height: 130px; width: 400px;"></a></p>
+<p><a href="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0160.Intersection%20of%20Two%20Linked%20Lists/images/160_example_1.png" target="_blank"><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0160.Intersection%20of%20Two%20Linked%20Lists/images/160_example_1.png" style="height: 130px; width: 400px;"></a></p>
 
 <pre><strong>输入：</strong>intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
 <strong>输出：</strong>Reference of the node with value = 8
@@ -29,7 +29,7 @@
 
 <p><strong>示例&nbsp;2：</strong></p>
 
-<p><a href="https://assets.leetcode.com/uploads/2018/12/13/160_example_2.png" target="_blank"><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0160.Intersection%20of%20Two%20Linked%20Lists/images/160_example_2.png" style="height: 136px; width: 350px;"></a></p>
+<p><a href="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0160.Intersection%20of%20Two%20Linked%20Lists/images/160_example_2.png" target="_blank"><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0160.Intersection%20of%20Two%20Linked%20Lists/images/160_example_2.png" style="height: 136px; width: 350px;"></a></p>
 
 <pre><strong>输入：</strong>intersectVal&nbsp;= 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
 <strong>输出：</strong>Reference of the node with value = 2
@@ -40,7 +40,7 @@
 
 <p><strong>示例&nbsp;3：</strong></p>
 
-<p><a href="https://assets.leetcode.com/uploads/2018/12/13/160_example_3.png" target="_blank"><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0160.Intersection%20of%20Two%20Linked%20Lists/images/160_example_3.png" style="height: 126px; width: 200px;"></a></p>
+<p><a href="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0160.Intersection%20of%20Two%20Linked%20Lists/images/160_example_3.png" target="_blank"><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0160.Intersection%20of%20Two%20Linked%20Lists/images/160_example_3.png" style="height: 126px; width: 200px;"></a></p>
 
 <pre><strong>输入：</strong>intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
 <strong>输出：</strong>null

solution/0400-0499/0430.Flatten a Multilevel Doubly Linked List/README.md

@@ -85,7 +85,6 @@
 	<li><code>1 &lt;= Node.val &lt;= 10^5</code></li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
@@ -97,15 +96,77 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+"""
+# 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':
+        def preorder(pre, cur):
+            if cur is None:
+                return pre
+            cur.prev = pre
+            pre.next = cur
+
+            t = cur.next
+            tail = preorder(cur, cur.child)
+            cur.child = None
+            return preorder(tail, t)
+            
+        if head is None:
+            return None
+        dummy = Node(0, None, head, None)
+        preorder(dummy, head)
+        dummy.next.prev = None
+        return dummy.next
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+/*
+// Definition for a Node.
+class Node {
+    public int val;
+    public Node prev;
+    public Node next;
+    public Node child;
+};
+*/
+
+class Solution {
+    public Node flatten(Node head) {
+        if (head == null) {
+            return null;
+        }
+        Node dummy = new Node();
+        dummy.next = head;
+        preorder(dummy, head);
+        dummy.next.prev = null;
+        return dummy.next;
+    }
+
+    private Node preorder(Node pre, Node cur) {
+        if (cur == null) {
+            return pre;
+        }
+        cur.prev = pre;
+        pre.next = cur;
+
+        Node t = cur.next;
+        Node tail = preorder(cur, cur.child);
+        cur.child = null;
+        return preorder(tail, t);
+    }
+}
 ```
 
 ### **...**

solution/0400-0499/0430.Flatten a Multilevel Doubly Linked List/README_EN.md

@@ -6,11 +6,8 @@
 
 <p>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.</p>
 
-
-
 <p>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.</p>
 
-
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
 
@@ -91,21 +88,82 @@ After flattening the multilevel linked list it becomes:
 	<li><code>1 &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
 </ul>
 
-
 ## Solutions
 
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
-
+"""
+# 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':
+        def preorder(pre, cur):
+            if cur is None:
+                return pre
+            cur.prev = pre
+            pre.next = cur
+
+            t = cur.next
+            tail = preorder(cur, cur.child)
+            cur.child = None
+            return preorder(tail, t)
+            
+        if head is None:
+            return None
+        dummy = Node(0, None, head, None)
+        preorder(dummy, head)
+        dummy.next.prev = None
+        return dummy.next
 ```
 
 ### **Java**
 
 ```java
-
+/*
+// Definition for a Node.
+class Node {
+    public int val;
+    public Node prev;
+    public Node next;
+    public Node child;
+};
+*/
+
+class Solution {
+    public Node flatten(Node head) {
+        if (head == null) {
+            return null;
+        }
+        Node dummy = new Node();
+        dummy.next = head;
+        preorder(dummy, head);
+        dummy.next.prev = null;
+        return dummy.next;
+    }
+
+    private Node preorder(Node pre, Node cur) {
+        if (cur == null) {
+            return pre;
+        }
+        cur.prev = pre;
+        pre.next = cur;
+
+        Node t = cur.next;
+        Node tail = preorder(cur, cur.child);
+        cur.child = null;
+        return preorder(tail, t);
+    }
+}
 ```
 
 ### **...**

solution/0400-0499/0430.Flatten a Multilevel Doubly Linked List/Solution.java

@@ -1,32 +1,35 @@
+/*
+// Definition for a Node.
+class Node {
+    public int val;
+    public Node prev;
+    public Node next;
+    public Node child;
+};
+*/
+
 class Solution {
     public Node flatten(Node head) {
         if (head == null) {
             return null;
         }
-        dfs(head);
-        head.prev = null;
-        return head;
+        Node dummy = new Node();
+        dummy.next = head;
+        preorder(dummy, head);
+        dummy.next.prev = null;
+        return dummy.next;
     }
 
-    private Node dfs(Node head) {
-        Node cur = head;
-        while (cur != null) {
-            head.prev = cur;
-            Node next = cur.next;
-            if (cur.child != null) {
-                Node h = dfs(cur.child);
-                cur.child = null;
-                Node t = h.prev;
-                cur.next = h;
-                h.prev = cur;
-                t.next = next;
-                if (next != null) {
-                    next.prev = t;
-                }
-                head.prev = t;
-            }
-            cur = next;
+    private Node preorder(Node pre, Node cur) {
+        if (cur == null) {
+            return pre;
         }
-        return head;
+        cur.prev = pre;
+        pre.next = cur;
+
+        Node t = cur.next;
+        Node tail = preorder(cur, cur.child);
+        cur.child = null;
+        return preorder(tail, t);
     }
-}
+}

solution/0400-0499/0430.Flatten a Multilevel Doubly Linked List/Solution.py

@@ -0,0 +1,29 @@
+"""
+# 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':
+        def preorder(pre, cur):
+            if cur is None:
+                return pre
+            cur.prev = pre
+            pre.next = cur
+
+            t = cur.next
+            tail = preorder(cur, cur.child)
+            cur.child = None
+            return preorder(tail, t)
+            
+        if head is None:
+            return None
+        dummy = Node(0, None, head, None)
+        preorder(dummy, head)
+        dummy.next.prev = None
+        return dummy.next