# LeetCode 117. Populating Next Right Pointers in Each Node II

## Description

https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

Given a binary tree

```struct Node {
int val;
Node *left;
Node *right;
Node *next;
}
```

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 `NULL`.

Initially, all next pointers are set to `NULL`.

• You may only use constant extra space.
• Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.

Example 1:

```Input: root = [1,2,3,4,5,null,7]
Output: [1,#,2,3,#,4,5,7,#]
Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.
```

Constraints:

• The number of nodes in the given tree is less than `6000`.
• `-100 <= node.val <= 100`

## Python Solution

``````"""
# 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 not root:
return None

queue = []
queue.append(root)

while queue:
size = len(queue)

prev = None
for i in range(size):
node = queue.pop(0)

if node.right:
queue.append(node.right)

if node.left:
queue.append(node.left)

node.next = prev
prev = node

return root``````
• Time Complexity: O(N).
• Space Complexity: O(N).