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feat: add solutions to lc problems: No.0116,0117
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Lines changed: 697 additions & 86 deletions

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‎README.md‎

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- [将二叉搜索树转换为单链表](/lcci/17.12.BiNode/README.md)
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- [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)
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- [二叉树的边界](/solution/0500-0599/0545.Boundary%20of%20Binary%20Tree/README.md)
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- [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)
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- [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)
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### 数学
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‎README_EN.md‎

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

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

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

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<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>
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<pre>
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struct Node {
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</pre>
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<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>
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<p>Initially, all next pointers are set to <code>NULL</code>.</p>
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<p>&nbsp;</p>
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<p><strong>Follow up:</strong></p>
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<ul>
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<li>You may only use constant extra space.</li>
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<li>Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.</li>
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</ul>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<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>
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<pre>
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<strong>Input:</strong> root = [1,2,3,4,5,6,7]
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li>The number of nodes in the given tree is less than <code>4096</code>.</li>
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<li><code>-1000 &lt;= node.val &lt;= 1000</code></li>
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### **Python3**
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```python
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"""
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# Definition for a Node.
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class Node:
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def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
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self.val = val
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self.left = left
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self.right = right
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self.next = next
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"""
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class Solution:
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def connect(self, root: 'Node') -> 'Node':
82+
if root is None or (root.left is None and root.right is None):
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return root
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q = collections.deque([root])
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while q:
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size = len(q)
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cur = None
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for _ in range(size):
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node = q.popleft()
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if node.right:
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q.append(node.right)
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if node.left:
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q.append(node.left)
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node.next = cur
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cur = node
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return root
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```
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### **Java**
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```java
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/*
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// Definition for a Node.
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class Node {
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public int val;
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public Node left;
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public Node right;
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public Node next;
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public Node() {}
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public Node(int _val) {
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val = _val;
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}
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public Node(int _val, Node _left, Node _right, Node _next) {
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val = _val;
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left = _left;
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right = _right;
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next = _next;
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}
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};
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*/
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class Solution {
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public Node connect(Node root) {
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if (root == null || (root.left == null && root.right == null)) {
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return root;
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}
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Deque<Node> q = new ArrayDeque<>();
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q.offer(root);
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while (!q.isEmpty()) {
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Node cur = null;
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for (int i = 0, n = q.size(); i < n; ++i) {
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Node node = q.pollFirst();
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if (node.right != null) {
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q.offer(node.right);
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}
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if (node.left != null) {
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q.offer(node.left);
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}
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node.next = cur;
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cur = node;
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}
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}
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return root;
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}
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}
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```
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### **C++**
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```cpp
154+
/*
155+
// Definition for a Node.
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class Node {
157+
public:
158+
int val;
159+
Node* left;
160+
Node* right;
161+
Node* next;
162+
163+
Node() : val(0), left(NULL), right(NULL), next(NULL) {}
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165+
Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}
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167+
Node(int _val, Node* _left, Node* _right, Node* _next)
168+
: val(_val), left(_left), right(_right), next(_next) {}
169+
};
170+
*/
171+
172+
class Solution {
173+
public:
174+
Node* connect(Node* root) {
175+
if (!root || (!root->left && !root->right)) {
176+
return root;
177+
}
178+
queue<Node*> q;
179+
q.push(root);
180+
while (!q.empty()) {
181+
Node* cur = nullptr;
182+
for (int i = 0, n = q.size(); i < n; ++i) {
183+
Node* node = q.front();
184+
q.pop();
185+
if (node->right) {
186+
q.push(node->right);
187+
}
188+
if (node->left) {
189+
q.push(node->left);
190+
}
191+
node->next = cur;
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cur = node;
193+
}
194+
}
195+
return root;
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}
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};
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```
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### **...**

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