117.PopulatingNextRightPointersinEachNodeII.py

"""
    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.

    Follow up:
        - 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                1 - NULL
                   / \              / \
                  2   3            2 - 3 - NULL
                 / \   \          / \ / \
                4  5    7        4--5----7 - NULL
    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
"""
#Difficulty: Medium
#55 / 55 test cases passed.
#Runtime: 60 ms
#Memory Usage: 15 MB

#Runtime: 60 ms, faster than 40.30% of Python3 online submissions for Populating Next Right Pointers in Each Node II.
#Memory Usage: 15 MB, less than 29.32% of Python3 online submissions for Populating Next Right Pointers in Each Node II.

"""
# 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
        queue = [root]
        while queue:
            length = len(queue)
            level = []
            while length:
                node = queue.pop(0)
                level.append(node)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
                length -= 1
            l = len(level)
            for i in range(l):
                if i == l - 1:
                    level[i].next = None
                if i + 1 < l:
                    level[i].next = level[i+1]
        return root