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Populating Next Right Pointers in Each Node

Problem Statement​

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

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.

Leetcode Link

Example 1:​

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 level.

Example 2:​

Input: root = []
Output: []

Constraints:​

  • The number of nodes in the tree is in the range [0, 212 - 1].
  • -1000 <= Node.val <= 1000

Code​

Python Code
from collections import deque
class Solution:
    # @param root, a tree link node
    # @return nothing
    def connect(self, root):
        if root is None: return
        q = deque([root])
        prevRight = None
        while len(q) > 0:
            cur = q.popleft()
            if cur.right is not None:
                if cur.next is not None:
                    cur.right.next = cur.next.left
                q.append(cur.right)
            if cur.left is not None:
                cur.left.next = cur.right
                q.append(cur.left)