116. Populating Next Right Pointers in Each Node

Problem Description

In this problem, we are dealing with a perfect binary tree where all leaves are at the same level, and every parent node has exactly two children. Each node in the tree has an additional pointer called next in addition to the usual left and right child pointers. Initially, all next pointers are set to NULL. The goal is to modify the tree such that each node’s next pointer points to the following node on the same level. For the nodes at the far right of each level, since there is no next right node, their next pointer should remain set to NULL.

Intuition

The solution approach is based on a level-order traversal (also known as a breadth-first search) of the tree. Since we're dealing with a perfect binary tree, every level is fully filled, and therefore the problem simplifies to connecting all nodes at each level in a left-to-right manner.

Here is the intuition step by step:

1. We initialize a queue that will store nodes at the current level we're processing.
2. We begin traversal with the root node by adding it into the queue.
3. We start a loop that will continue until the queue is empty, which means there are no more levels to process.
4. Inside this loop, we initialize a variable p to None, to keep track of the previous node as we process.
5. We then process nodes level by level:
   • At the start of a level, we record the number of nodes at that level (the queue’s length) to ensure we only process those nodes before moving to the next level.
   • We pop nodes off of the queue left to right, and for each node: a. If p is not NULL (this is not the first node of the level), we set p's next to the current node. b. We then update p to the current node. c. We add the current node's left and right children to the queue.
6. By the end of the loop, all next pointers should accurately point to the next right node, or remain NULL if it's the far-right node of that level.
7. Finally, we return the modified tree starting with the root node.

The given solution ensures that each node at a given level points to its adjacent node on the right before descending to the next level. It takes advantage of the queue data structure for keeping track of nodes level by level and updates the next pointers in a systematic manner.

Learn more about Tree, Depth-First Search, Breadth-First Search, Linked List and Binary Tree patterns.

Solution Approach

The implementation of the solution is primarily based on the Breadth-First Search (BFS) algorithm, which is a common way to traverse trees level by level. To make this possible, we employ a queue data structure, which will hold the nodes of the tree according to the BFS traversal order. The Python deque from the collections module is used for the queue because it allows for fast and efficient appending and popping of elements from both ends.

Here is a breakdown of the implementation steps, referring to the data structures and patterns used:

1. We first check if the root is None. If it is, we return None as there is nothing to connect.
2. We initialize a queue q with the root node of the tree.
3. We enter a while loop that will run as long as there are elements in the queue.
4. In each iteration of the while loop, we first initialize a variable p to None, which will serve as a pointer to the previous node at the current level.
5. We determine the number of nodes at the current level of the tree by examining the length of the queue using len(q).
6. We then use a for loop to iterate over all nodes at the current level:
   • We use q.popleft() to remove and obtain the first node from the queue.
   • We check if p is not None. This implies that this is not the first node at the current level, so we set the next of p to the current node.
   • We set p to the current node after updating its next.
   • If the current node has a left child, we append it to the queue, and similarly, if it has a right child, we append it to the queue.
7. Finally, once the while loop finishes executing, all nodes' next pointers have been properly set, and we return the root of the updated tree.

This solution utilizes the properties of a perfect binary tree and a level-order traversal algorithm to connect the nodes at each level efficiently. With the help of the queue, we are able to process nodes in the correct order and update their next pointers while traversing the tree.

Here's a snippet of the key part of the implementation that demonstrates the core process:

1while q:
2    p = None
3    for _ in range(len(q)):
4        node = q.popleft()
5        if p:
6            p.next = node
7        p = node
8        if node.left:
9            q.append(node.left)
10        if node.right:
11            q.append(node.right)

In this snippet, the loop structures and the assignments are critical in understanding how the connections (based on next pointers) are made between nodes at the same level.

Example Walkthrough

Let's consider a small example of a perfect binary tree to illustrate the solution approach. Our example tree is as follows (represented as levels):

1    1
2   / \
3  2   3
4 / \ / \
54  5 6  7

To begin, here is an outline of the steps we’ll follow to execute the algorithm described:

1. Create an empty queue (will use a deque for efficiency) and enqueue the root node (node 1).
2. Process the nodes until the queue is empty.

Now let's walkthrough each level:

• Level 0: Tree root (Node 1)

  • Initialize queue as deque([1]).
  • p = None
  • There is only one node at this level, so we don't have to set any next pointers.
  • Enqueue its children, queue becomes deque([2, 3]).

• Level 1: Nodes 2 and 3

  • p = None, we then dequeue node 2 and since p is None, next is not set.
  • p is updated to node 2. Node 2’s children (4, 5) are enqueued, queue becomes deque([3, 4, 5]).
  • We dequeue node 3, and set next of node 2 to node 3 (p.next = node), p is updated to node 3.
  • Node 3's children (6, 7) are enqueued, queue becomes deque([4, 5, 6, 7]).

• Level 2: Nodes 4, 5, 6, and 7

  • p = None, we then dequeue node 4 and since p is None, next is not set.
  • p is updated to node 4.
  • Continue with nodes 5, 6, and 7 in the same way, dequeuing each node, setting its next pointer to the subsequent node, and updating p to the latest node.
  • After node 7, the queue is empty, and the next pointer is not updated as it is the last node at this level.

At the end of this process, here’s how the next pointers would be set:

1    1 -> NULL
2   / \
3  2 -> 3 -> NULL
4 / \ / \
54->5->6->7 -> NULL

All the next pointers at the same level have been connected, and the next pointers of the last nodes at each level remain pointing to NULL, signifying the end of the level. This completes the walk through of the tree using the given solution approach.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code is O(N), where N is the total number of nodes in the binary tree. The algorithm employs a breadth-first search approach, where each node is visited exactly once. Because we visit each node once and perform a constant amount of work for each node (like setting the next pointers and pushing children to the queue), the overall time complexity is linear with respect to the number of nodes.

Space Complexity

The space complexity of the code is O(N). The space is mainly occupied by the queue q, which in the worst-case scenario can contain all nodes at the deepest level of the binary tree. In a perfect binary tree, the last level could contain approximately N/2 nodes, where N is the total number of nodes. Thus, the space complexity is proportional to the number of nodes in the bottom level of the tree, making it O(N) in the worst case.

Learn more about how to find time and space complexity quickly using problem constraints.