117. Populating Next Right Pointers in Each Node II

Medium

---

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.

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.

Example 2:

Input: root = []
Output: []

Constraints:

• The number of nodes in the tree is in the range [0, 6000].

• -100 <= Node.val <= 100

Follow-up:

• You may only use constant extra space.

• The recursive approach is fine. You may assume implicit stack space does not count as extra space for this problem.

Solution 1 :

class Solution:
    def connect(self, root: 'Node') -> 'Node':
        if not root:
            return root
        dq = deque()
        dq.append(root)
        dq.append('#')
        prev = None
        while dq:
            node = dq.popleft()
            if node == '#':
                if prev:
                    prev.next = None
                    prev = None
                if len(dq) != 0:
                    dq.append('#')
                continue
            if node.left:
                dq.append(node.left)
            if node.right:
                dq.append(node.right)
            if prev:
                prev.next = node
            prev = node
        return root

"""
# 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
        q = [root,'#']
        prev = None
        while q:
            node = q.pop(0)
            if node == '#':
                if prev is not None:
                    prev.next = None
                if q:
                    # node = q.pop(0)
                    prev = None
                    q.append('#')
                    continue
                else:
                    break
            else:
                if prev is not None:
                    prev.next = node
                prev = node
            if node.left:
                q.append(node.left)
            if node.right:
                q.append(node.right)
        return root
            
            
pp