Populating Next Right Pointers In Each Node

Tested Python solution for LeetCode 116 with 16 pytest cases. Generate a practice environment with lcpy.

LeetCode 116, Medium. Topics: Linked List, Tree, Depth-First Search, Breadth-First Search, Binary Tree. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 16 parametrized pytest cases, and a playground notebook:

lcpy gen -n 116   # by problem number
lcpy gen -s populating_next_right_pointers_in_each_node   # by problem name

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Problem

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.

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Examples

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.

Input: root = []
Output: []

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Constraints

• The number of nodes in the tree is in the range [0, 2^12 - 1]
• -1000 <= Node.val <= 1000

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.

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

from __future__ import annotations


class Node:
    def __init__(
        self,
        val: int = 0,
        left: Node | None = None,
        right: Node | None = None,
        next: Node | None = None,
    ):
        self.val = val
        self.left = left
        self.right = right
        self.next = next


class Solution:
    # Time: O(n)
    # Space: O(1)
    def connect(self, root: Node | None) -> Node | None:
        leftmost = root
        while leftmost is not None and leftmost.left is not None:
            head = leftmost
            while head is not None:
                left = head.left
                right = head.right
                assert left is not None and right is not None
                left.next = right
                if head.next is not None:
                    right.next = head.next.left
                head = head.next
            leftmost = leftmost.left
        return root

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Complexity

Time	Space
O(n)	O(1)

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Tags

NeetCode All.