feat: add solutions to lc problems: No.0116,0117

README.md

@@ -126,6 +126,8 @@
 - [将二叉搜索树转换为单链表](/lcci/17.12.BiNode/README.md)
 - [将二叉搜索树转化为排序的双向链表](/solution/0400-0499/0426.Convert%20Binary%20Search%20Tree%20to%20Sorted%20Doubly%20Linked%20List/README.md)
 - [二叉树的边界](/solution/0500-0599/0545.Boundary%20of%20Binary%20Tree/README.md)
+- [填充每个节点的下一个右侧节点指针](/solution/0100-0199/0116.Populating%20Next%20Right%20Pointers%20in%20Each%20Node/README.md)
+- [填充每个节点的下一个右侧节点指针 II](/solution/0100-0199/0117.Populating%20Next%20Right%20Pointers%20in%20Each%20Node%20II/README.md)
 
 ### 数学
 

README_EN.md

@@ -122,6 +122,8 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
 - [BiNode](/lcci/17.12.BiNode/README_EN.md)
 - [Convert Binary Search Tree to Sorted Doubly Linked List](/solution/0400-0499/0426.Convert%20Binary%20Search%20Tree%20to%20Sorted%20Doubly%20Linked%20List/README_EN.md)
 - [Boundary of Binary Tree](/solution/0500-0599/0545.Boundary%20of%20Binary%20Tree/README_EN.md)
+- [Populating Next Right Pointers in Each Node](/solution/0100-0199/0116.Populating%20Next%20Right%20Pointers%20in%20Each%20Node/README_EN.md)
+- [Populating Next Right Pointers in Each Node II](/solution/0100-0199/0117.Populating%20Next%20Right%20Pointers%20in%20Each%20Node%20II/README_EN.md)
 
 ### Math
 

solution/0100-0199/0116.Populating Next Right Pointers in Each Node/README.md

@@ -50,27 +50,149 @@ struct Node {
 	<li><code>-1000 <= node.val <= 1000</code></li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+“BFS 层次遍历”实现。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 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: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
+        self.val = val
+        self.left = left
+        self.right = right
+        self.next = next
+"""
+
+class Solution:
+    def connect(self, root: 'Node') -> 'Node':
+        if root is None or (root.left is None and root.right is None):
+            return root
+        q = collections.deque([root])
+        while q:
+            size = len(q)
+            cur = None
+            for _ in range(size):
+                node = q.popleft()
+                if node.right:
+                    q.append(node.right)
+                if node.left:
+                    q.append(node.left)
+                node.next = cur
+                cur = node
+        return root
 ```
 
 ### **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 left;
+    public Node right;
+    public Node next;
+
+    public Node() {}
+
+    public Node(int _val) {
+        val = _val;
+    }
+
+    public Node(int _val, Node _left, Node _right, Node _next) {
+        val = _val;
+        left = _left;
+        right = _right;
+        next = _next;
+    }
+};
+*/
+
+class Solution {
+    public Node connect(Node root) {
+        if (root == null || (root.left == null && root.right == null)) {
+            return root;
+        }
+        Deque<Node> q = new ArrayDeque<>();
+        q.offer(root);
+        while (!q.isEmpty()) {
+            Node cur = null;
+            for (int i = 0, n = q.size(); i < n; ++i) {
+                Node node = q.pollFirst();
+                if (node.right != null) {
+                    q.offer(node.right);
+                }
+                if (node.left != null) {
+                    q.offer(node.left);
+                }
+                node.next = cur;
+                cur = node;
+            }
+        }
+        return root;
+    }
+}
+```
 
+### **C++**
+
+```cpp
+/*
+// Definition for a Node.
+class Node {
+public:
+    int val;
+    Node* left;
+    Node* right;
+    Node* next;
+
+    Node() : val(0), left(NULL), right(NULL), next(NULL) {}
+
+    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}
+
+    Node(int _val, Node* _left, Node* _right, Node* _next)
+        : val(_val), left(_left), right(_right), next(_next) {}
+};
+*/
+
+class Solution {
+public:
+    Node* connect(Node* root) {
+        if (!root || (!root->left && !root->right)) {
+            return root;
+        }
+        queue<Node*> q;
+        q.push(root);
+        while (!q.empty()) {
+            Node* cur = nullptr;
+            for (int i = 0, n = q.size(); i < n; ++i) {
+                Node* node = q.front();
+                q.pop();
+                if (node->right) {
+                    q.push(node->right);
+                }
+                if (node->left) {
+                    q.push(node->left);
+                }
+                node->next = cur;
+                cur = node;
+            }
+        }
+        return root;
+    }
+};
 ```
 
 ### **...**

solution/0100-0199/0116.Populating Next Right Pointers in Each Node/README_EN.md

@@ -6,8 +6,6 @@
 
 <p>You are given a <strong>perfect binary tree</strong>&nbsp;where&nbsp;all leaves are on the same level, and every parent has two children. The binary tree has the following definition:</p>
 
-
-
 <pre>
 
 struct Node {
@@ -24,41 +22,25 @@ struct Node {
 
 </pre>
 
-
-
 <p>Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to <code>NULL</code>.</p>
 
-
-
 <p>Initially, all next pointers are set to <code>NULL</code>.</p>
 
-
-
 <p>&nbsp;</p>
 
-
-
 <p><strong>Follow up:</strong></p>
 
-
-
 <ul>
 	<li>You may only use constant extra space.</li>
 	<li>Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.</li>
 </ul>
 
-
-
 <p>&nbsp;</p>
 
 <p><strong>Example 1:</strong></p>
 
-
-
 <p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0116.Populating%20Next%20Right%20Pointers%20in%20Each%20Node/images/116_sample.png" style="width: 640px; height: 218px;" /></p>
 
-
-
 <pre>
 
 <strong>Input:</strong> root = [1,2,3,4,5,6,7]
@@ -69,14 +51,10 @@ struct Node {
 
 </pre>
 
-
-
 <p>&nbsp;</p>
 
 <p><strong>Constraints:</strong></p>
 
-
-
 <ul>
 	<li>The number of nodes in the given tree is less than <code>4096</code>.</li>
 	<li><code>-1000 &lt;= node.val &lt;= 1000</code></li>
@@ -89,13 +67,134 @@ struct Node {
 ### **Python3**
 
 ```python
-
+"""
+# Definition for a Node.
+class Node:
+    def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
+        self.val = val
+        self.left = left
+        self.right = right
+        self.next = next
+"""
+
+class Solution:
+    def connect(self, root: 'Node') -> 'Node':
+        if root is None or (root.left is None and root.right is None):
+            return root
+        q = collections.deque([root])
+        while q:
+            size = len(q)
+            cur = None
+            for _ in range(size):
+                node = q.popleft()
+                if node.right:
+                    q.append(node.right)
+                if node.left:
+                    q.append(node.left)
+                node.next = cur
+                cur = node
+        return root
 ```
 
 ### **Java**
 
 ```java
+/*
+// Definition for a Node.
+class Node {
+    public int val;
+    public Node left;
+    public Node right;
+    public Node next;
+
+    public Node() {}
+
+    public Node(int _val) {
+        val = _val;
+    }
+
+    public Node(int _val, Node _left, Node _right, Node _next) {
+        val = _val;
+        left = _left;
+        right = _right;
+        next = _next;
+    }
+};
+*/
+
+class Solution {
+    public Node connect(Node root) {
+        if (root == null || (root.left == null && root.right == null)) {
+            return root;
+        }
+        Deque<Node> q = new ArrayDeque<>();
+        q.offer(root);
+        while (!q.isEmpty()) {
+            Node cur = null;
+            for (int i = 0, n = q.size(); i < n; ++i) {
+                Node node = q.pollFirst();
+                if (node.right != null) {
+                    q.offer(node.right);
+                }
+                if (node.left != null) {
+                    q.offer(node.left);
+                }
+                node.next = cur;
+                cur = node;
+            }
+        }
+        return root;
+    }
+}
+```
 
+### **C++**
+
+```cpp
+/*
+// Definition for a Node.
+class Node {
+public:
+    int val;
+    Node* left;
+    Node* right;
+    Node* next;
+
+    Node() : val(0), left(NULL), right(NULL), next(NULL) {}
+
+    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}
+
+    Node(int _val, Node* _left, Node* _right, Node* _next)
+        : val(_val), left(_left), right(_right), next(_next) {}
+};
+*/
+
+class Solution {
+public:
+    Node* connect(Node* root) {
+        if (!root || (!root->left && !root->right)) {
+            return root;
+        }
+        queue<Node*> q;
+        q.push(root);
+        while (!q.empty()) {
+            Node* cur = nullptr;
+            for (int i = 0, n = q.size(); i < n; ++i) {
+                Node* node = q.front();
+                q.pop();
+                if (node->right) {
+                    q.push(node->right);
+                }
+                if (node->left) {
+                    q.push(node->left);
+                }
+                node->next = cur;
+                cur = node;
+            }
+        }
+        return root;
+    }
+};
 ```
 
 ### **...**