<p>Given a binary tree</p>

<pre>
struct Node {
  int val;
  Node *left;
  Node *right;
  Node *next;
}
</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 class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/02/15/117_sample.png" style="width: 500px; height: 171px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4,5,null,7]
<strong>Output:</strong> [1,#,2,3,#,4,5,7,#]
<strong>Explanation: </strong>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 &#39;#&#39; signifying the end of each level.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = []
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 6000]</code>.</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow-up:</strong></p>

<ul>
	<li>You may only use constant extra space.</li>
	<li>The recursive approach is fine. You may assume implicit stack space does not count as extra space for this problem.</li>
</ul>

# 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 root
        node = root
        next_node = None
        while node:
            if node.left:
                if not next_node:
                    next_node = node.left
            elif node.right:
                if not next_node:
                    next_node = node.right

            if node.left:
                if node.right:
                    node.left.next = node.right
                else:
                    nd = node.next
                    while nd:
                        if nd.left:
                            node.left.next = nd.left
                            break
                        elif nd.right:
                            node.left.next = nd.right
                            break
                        else:
                            pass
                        nd = nd.next
                if next_node and not next_node.next and next_node != node.left:
                    next_node.next = node.left
            if node.right:
                nd = node.next
                while nd:
                    if nd.left:
                        node.right.next = nd.left
                        break
                    elif nd.right:
                        node.right.next = nd.right
                        break
                    else:
                        pass
                    nd = nd.next
                if next_node and not next_node.next and next_node != node.right:
                    next_node.next = node.right
            if node.next:
                node = node.next
            else:
                node = next_node
                next_node = None
        return root
17700