Populating Next Right Pointers In Each Node Ii
To solve this coding challenge, we need to populate the
pointers of a binary tree such that each node's
pointer points to its next right node. If there is no next right node, the
pointer should be set to
. Let's break this problem down step-by-step and generate the pseudocode for the solution.
next
next
next
NULL
Explanation
- Understanding the Tree Structure :
-
Each node of the tree has a
val
left
right
next
-
Our task is to link the
next
next
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
- Add the children of each node to the queue to process the next level.
- Edge Cases :
-
An empty tree should return
[]
-
Nodes without children should have their
next
NULL
- Node and Tree Definition :
-
The
Node
val
left
right
next
-
The
connect
- Initial Root Check :
-
If the root is
None
None
- 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
-
If it is the last node, set its
next
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
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