# [117. 填充每个节点的下一个右侧节点指针 II](https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node-ii)
[English Version](/solution/0100-0199/0117.Populating%20Next%20Right%20Pointers%20in%20Each%20Node%20II/README_EN.md)
## 题目描述
给定一个二叉树
struct Node {
int val;
Node *left;
Node *right;
Node *next;
}
填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。
初始状态下，所有 next 指针都被设置为 NULL。
进阶：
- 你只能使用常量级额外空间。
- 使用递归解题也符合要求，本题中递归程序占用的栈空间不算做额外的空间复杂度。
示例：
输入：root = [1,2,3,4,5,null,7]
输出：[1,#,2,3,#,4,5,7,#]
解释：给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。序列化输出按层序遍历顺序（由 next 指针连接），'#' 表示每层的末尾。
提示：
- 树中的节点数小于
6000
-100 <= node.val <= 100
## 解法
“BFS 层次遍历”实现。
### **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 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 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;
}
};
```
### **...**
```
```