LeetCode 116: Populating Next Right Pointers in Each Node Solution in Python – A Step-by-Step Guide

What Is LeetCode 116: Populating Next Right Pointers in Each Node?

In LeetCode 116: Populating Next Right Pointers in Each Node, you’re given the root of a perfect binary tree—every node has either 0 or 2 children, and all leaves are at the same depth. Each node has a val, left, right, and next pointer (initially None). Your task is to populate each next pointer to point to the node immediately to its right at the same level, or None if it’s the last node in the level. For example, a tree like [1, 2, 3, 4, 5, 6, 7] becomes [1,#, 2,3,#, 4,5,6,7,#], where # marks the end of a level.

Problem Statement

• Input: Root of a perfect binary tree (Node).
• Output: None (modify the tree in-place).
• Rules:
• Set each node’s next to the right neighbor at the same level.
• Use constant extra space (O(1), excluding recursion stack).
• Tree is perfect—all levels fully filled except possibly the last.

Constraints

• Number of nodes: 0 <= n <= 2¹² - 1 (up to 4095).
• -1000 <= Node.val <= 1000

Examples

• Input: root = [1, 2, 3, 4, 5, 6, 7]
• Output: [1,#, 2,3,#, 4,5,6,7,#]
• Why: 2→3, 4→5→6→7.
• Input: root = []
• Output: []
• Why: Empty tree, nothing to connect.
• Input: root = [1]
• Output: [1,#]
• Why: Single node, next is None.

Understanding the Problem: Linking the Levels

To solve LeetCode 116: Populating Next Right Pointers in Each Node in Python, we need to connect nodes at the same level in a perfect binary tree using the next pointer, all while keeping extra space to O(1). A naive approach might use a queue for level-order traversal (O(n) space), but the “constant space” hint and perfect tree structure suggest we can leverage the tree itself. We’ll explore:

• Best Solution (Level-Order with O(1) Space): O(n) time, O(1) space—uses next pointers to traverse.
• Alternative Solution (Recursive with O(1) Space): O(n) time, O(1) space—elegant recursion.

Let’s dive into the Best Solution—level-order with next pointers—and break it down clearly.

Best Solution: Level-Order Traversal with O(1) Space

Why This Is the Best Solution

The level-order traversal using next pointers is the top pick because it achieves O(n) time and O(1) space without recursion, making it both efficient and practical. It uses the next pointers as they’re being set to navigate each level, turning the tree’s own structure into a makeshift queue. It’s like laying down a rope as you climb, then using it to swing across each level—ingenious and space-savvy!

How It Works

Here’s the strategy:

• Step 1: Start at the Root:
• Begin with the leftmost node of the top level (root).
• Step 2: Connect a Level:
• While a level has nodes:
• Link each node’s left child to its right child.
• Link right child to the left child of the next node (via next).
• Step 3: Move Down:
• Use the next pointers to move to the leftmost node of the next level.
• Step 4: Repeat:
• Continue until no more levels with children.

Step-by-Step Example

Example: root = [1, 2, 3, 4, 5, 6, 7]

• Initial Tree:

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

• Level 0 (Root):
• curr = 1, no siblings to connect, move to left child.
• Level 1 (2, 3):
• curr = 2:
• 4.next = 5 (left to right).
• 5.next = 6 (right to next node’s left, via 2.next = 3).
• curr = 3:
• 6.next = 7.
• 7.next = None (end of level).
• Tree:

1 / \ 2→3 / \ / \ 4→5→6→7

• Level 2 (4, 5, 6, 7):
• No children, done.
• Result: [1,#, 2,3,#, 4,5,6,7,#].

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained step-by-step:

Copy codeclass Solution:
    def connect(self, root: 'Node') -> 'Node':
        # Handle empty tree
        if not root:
            return None

        # Start at root
        leftmost = root

        # Process each level
        while leftmost.left:  # While there are children
            curr = leftmost
            # Connect nodes in current level
            while curr:
                # Connect left child to right child
                curr.left.next = curr.right
                # Connect right child to next node's left child
                if curr.next:
                    curr.right.next = curr.next.left
                # Move to next node in level
                curr = curr.next
            # Move to next level
            leftmost = leftmost.left

        return root

• Line 4-5: Return None for empty tree.
• Line 8: Start with leftmost at root.
• Line 11-19: Loop over levels:
• Line 12: curr tracks nodes in current level.
• Line 15: Link left to right child.
• Line 17-18: Link right to next node’s left (if exists).
• Line 20: Move to next node in level.
• Line 22: Move to next level via leftmost child.
• Time Complexity: O(n)—visit each node once.
• Space Complexity: O(1)—no extra data structures.

This weaves the tree into a connected web!

Alternative Solution: Recursive Approach with O(1) Space

Why an Alternative Approach?

The recursive approach offers a clean, top-down way to connect nodes—O(n) time, O(1) space (excluding recursion stack, which problem allows). It uses the perfect tree property to link children recursively, making it elegant and concise. It’s like telling each parent to sort out their kids, then passing the job down—simple and clever!

How It Works

Here’s the plan:

• Step 1: Base case—empty node, do nothing.
• Step 2: Connect children of current node:
• Left.next = Right.
• Right.next = Parent.next’s left (if parent has a next).
• Step 3: Recurse on left and right subtrees.

Step-by-Step Example

Example: root = [1, 2, 3, 4, 5, 6, 7]

• Node 1:
• 2.next = 3.
• Recurse on 2 and 3.
• Node 2:
• 4.next = 5.
• 5.next = 6 (via 2.next = 3).
• Recurse on 4 and 5.
• Node 3:
• 6.next = 7.
• 7.next = None.
• Recurse on 6 and 7.
• Result: [1,#, 2,3,#, 4,5,6,7,#].

Code for Recursive Approach

Copy codeclass Solution:
    def connect(self, root: 'Node') -> 'Node':
        # Base case
        if not root:
            return None

        # Connect children
        if root.left:
            root.left.next = root.right
            if root.next:
                root.right.next = root.next.left

        # Recurse on subtrees
        self.connect(root.left)
        self.connect(root.right)

        return root

• Line 4-5: Handle empty tree.
• Line 8-11: Connect children:
• Line 9: Left to right.
• Line 11: Right to next level’s left (if exists).
• Line 14-15: Recurse on both subtrees.
• Time Complexity: O(n)—each node processed once.
• Space Complexity: O(1) extra space (recursion stack excluded).

It’s a recursive connector!

Comparing the Two Solutions

• Level-Order (Best):
• Pros: O(n) time, O(1) space, iterative, explicit level control.
• Cons: Slightly more code.
• Recursive (Alternative):
• Pros: O(n) time, O(1) space, concise, elegant.
• Cons: Relies on recursion stack (still meets problem’s O(1) intent).

Level-order wins for clarity and iteration.

Additional Examples and Edge Cases

• [1, 2, 3]: [1,#, 2,3,#].
• [1, 2, 3, 4, 5]: [1,#, 2,3,#, 4,5,#].
• []: [].

Both handle these perfectly.

Complexity Breakdown

• Level-Order: Time O(n), Space O(1).
• Recursive: Time O(n), Space O(1) (excluding stack).

Level-order’s iterative edge shines.

Key Takeaways

• Level-Order: Use next to traverse!
• Recursive: Connect and descend!
• Perfect Tree: Symmetry helps.
• Python Tip: Pointers simplify—see [Python Basics](/python/basics).

Final Thoughts: Link the Levels

LeetCode 116: Populating Next Right Pointers in Each Node in Python is a tree-linking adventure. The level-order approach weaves connections with finesse, while recursion offers a sleek alternative. Want more tree challenges? Try LeetCode 117: Populating Next Right Pointers II or LeetCode 102: Binary Tree Level Order Traversal. Ready to connect? Head to Solve LeetCode 116 on LeetCode and link those nodes today!