Populating Next Right Pointers In Each Node Ii

                                            
# Definition for a Node.
class Node(object):
    def __init__(self, val=0, left=None, right=None, next=None):
        self.val = val
        self.left = left
        self.right = right
        self.next = next

class Solution(object):
    def connect(self, root):
        if not root:
            return None
        queue = [root]
        while queue:
            size = len(queue)
            for i in range(size):
                node = queue.pop(0)
                if i < size - 1:
                    node.next = queue[0]
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
        return root

                                            
/**
 * // Definition for a Node.
 * function Node(val, left, right, next) {
 *    this.val = val === undefined ? null : val;
 *    this.left = left === undefined ? null : left;
 *    this.right = right === undefined ? null : right;
 *    this.next = next === undefined ? null : next;
 * }
 */

var connect = function(root) {
    if (!root) return root;
    
    let queue = [root];
    
    while (queue.length > 0) {
        let levelSize = queue.length;
        for (let i = 0; i < levelSize; i++) {
            let currentNode = queue.shift();
            if (i < levelSize - 1) {
                currentNode.next = queue[0];
            }
            if (currentNode.left) {
                queue.push(currentNode.left);
            }
            if (currentNode.right) {
                queue.push(currentNode.right);
            }
        }
    }
    
    return root;
};

                                            
/**
 * Definition for Node.
 * class Node {
 *     val: number
 *     left: Node | null
 *     right: Node | null
 *     next: Node | null
 * }
 */
function connect(root: Node | null): Node | null {
    if (!root) return root;
    
    let queue: Node[] = [root];
    
    while (queue.length > 0) {
        let size = queue.length;
        
        for (let i = 0; i < size; i++) {
            let node = queue.shift()!;
            
            if (i < size - 1) {
                node.next = queue[0];
            }
            
            if (node.left) queue.push(node.left);
            if (node.right) queue.push(node.right);
        }
    }
    
    return root;
}

                                            
/**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Left *Node
 *     Right *Node
 *     Next *Node
 * }
 */
func connect(root *Node) *Node {
    if root == nil {
        return root
    }
    
    queue := []*Node{root}
    
    for len(queue) > 0 {
        size := len(queue)
        
        for i := 0; i < size; i++ {
            node := queue[0]
            queue = queue[1:]
            
            if i < size-1 {
                node.Next = queue[0]
            }
            
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
        }
    }
    
    return root
}

                                            
# Definition for a Node.
# class Node
#     attr_accessor :val, :left, :right, :next
#     def initialize(val)
#         @val = val
#         @left = nil
#         @right = nil
#         @next = nil
#     end
# end

def connect(root)
    return nil if root.nil?
    
    queue = [root]
    
    while !queue.empty?
        level_size = queue.size
        
        level_size.times do |i|
            current = queue.shift
            queue << current.left if current.left
            queue << current.right if current.right
            
            current.next = queue.first if i < level_size - 1
        end
    end
    
    return root
end

                                            
/**
 * Definition for a Node.
 * class Node(var _value: Int) {
 *   var value: Int = _value
 *   var left: Node = null
 *   var right: Node = null
 *   var next: Node = null
 * }
 */

object Solution {
  def connect(root: Node): Node = {
    if (root == null) return null

    val queue = scala.collection.mutable.Queue[Node]()
    queue.enqueue(root)

    while (queue.nonEmpty) {
      val size = queue.size
      var prev: Node = null
      for (_ <- 0 until size) {
        val current = queue.dequeue()
        if (prev != null) prev.next = current
        prev = current

        if (current.left != null) queue.enqueue(current.left)
        if (current.right != null) queue.enqueue(current.right)
      }
    }

    root
  }
}

To solve this coding challenge, we need to populate the

next

pointers of a binary tree such that each node's

next

pointer points to its next right node. If there is no next right node, the

next

pointer should be set to

NULL

. Let's break this problem down step-by-step and generate the pseudocode for the solution.

Explanation

Understanding the Tree Structure :

Each node of the tree has a
val

,
left

(left child),
right

(right child), and
next

(next right node) pointer.
Our task is to link the
next

pointer of each node to the node that lies to its immediate right at the same level in the binary tree. If no such node exists, the
next

pointer should be set to
NULL

.

Initial Setup :

If the tree is empty, return
None

.
If it’s not empty, we will use a breadth-first search (BFS) traversal technique to process nodes level by level. BFS works well for this task because it processes all nodes at the same level before moving on to the next level.

Using a Queue for BFS :

We will use a queue to keep track of nodes at each level.
Initialize the queue with the root node.

Processing Each Level :

For each level in the tree, process each node in the queue.
Link each node's
next

pointer to the node that follows it in the queue (if any).
Add the children of each node to the queue to process the next level.

Edge Cases :

An empty tree should return
[]

(empty list).
Nodes without children should have their
next

set to
NULL

.

Now, let's provide pseudocode to elaborate on these steps:

Pseudocode with Comments

                                            
# Begin by defining the structure of a Node and a placeholder for the tree root

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

# The solution function to connect the `next` pointers

def connect(root):
# If the tree root is `None`, return `None`
if root is None:
return None

# Initialize a queue for BFS with the root node
queue = [root]

# Loop until the queue is empty
while queue:

# Get the number of nodes at the current level
level_size = len(queue)

# Process all nodes at the current level
for i in range(level_size):

# Pop the first node from the queue
current_node = queue.pop(0)

# If this is not the last node in the level, set its `next` to the node in queue front
if i < level_size - 1:
current_node.next = queue[0]
else:
# If this is the last node, ensure its `next` is `NULL`
current_node.next = None

# Add the left child to the queue if it exists
if current_node.left is not None:
queue.append(current_node.left)

# Add the right child to the queue if it exists
if current_node.right is not None:
queue.append(current_node.right)

# Return the root after fully populating `next` pointers
return root

Step-by-Step Explanation of the Pseudocode

Node and Tree Definition :

The
Node

class is defined with
val

,
left

,
right

, and
next

attributes.
The
connect

function is defined to operate on the tree's root.

Initial Root Check :

If the root is
None

, it returns
None

immediately as there are no nodes to process.

Queue Initialization :

A queue is initialized with the root node to start BFS.

Processing Loop :

While the queue contains nodes, continue processing.

Level Processing :

Determine the size of the current level by the queue's length.
Iterate through each node in the current level.

Next Pointer Assignment :

For each node, set its
next

to the next node in the queue if it is not the last node in the level.
If it is the last node, set its
next

to
NULL

.

Child Nodes Enqueue :

Append the left and right children (if they exist) to the queue for the next level of processing.

Return the Root :

After all levels are processed and
next

pointers are set, return the root node of the modified tree.

This pseudocode outlines a clear and systematic approach to solving the problem efficiently using BFS, ensuring constant extra space usage aside from the implicit stack space.