Leetcode 116 Populating Next Right Pointers in Each Node
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.
Input: root = [1,2,3,4,5,6,7]
Output: [1,#,2,3,#,4,5,6,7,#]
Explanation: Given the above perfect 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
Input: root = []
Output: []

This question can be solved by Depth First Search

The tree is a perfect binary tree. So if there is left node then there must be right node. For each node, if it has left child node then we use next pointer to connect the left child node to right child node. If the node has next pointer, then the node’s right child node will use next pointer to connect to the node’s next node’s left child node.
"""
# 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:
return root
root.left.next = root.right
if root.next:
root.right.next = root.next.left
self.connect(root.left)
self.connect(root.right)
return root