116. Populating Next Right Pointers in Each Node - Detailed Explanation

Problem Statement

Description:
You are given a perfect binary tree where all leaves are on the same level and every parent has two children. The tree node is defined as follows:

class Node {
    public int val;
    public Node left;
    public Node right;
    public 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.

Examples:

• Example 1:

  Input:
  A perfect binary tree:

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

  Output:
  After connecting nodes, the tree should be modified as:

           1 -> null
         /   \
        2  -> 3 -> null
       / \   / \
      4->5->6->7 -> null
  

  Explanation:

  • Level 1: Node 1’s next remains null.
  • Level 2: Node 2’s next is Node 3; Node 3’s next is null.
  • Level 3: Node 4’s next is Node 5, Node 5’s next is Node 6, Node 6’s next is Node 7, and Node 7’s next is null.

Constraints:

• The number of nodes in the tree is in the range [0, 2^12 - 1].
• The tree is perfect, meaning all leaves are at the same level, and every parent has two children.

Hints

• Hint 1:
  Since the tree is perfect, you can use the fact that a node’s left child is always connected to its right child. Additionally, if the current node has a next pointer, then the current node’s right child should connect to the current node’s next’s left child.

• Hint 2:
  Think about a level-by-level traversal without using extra space. Can you use the next pointers that you create at one level to help traverse the next level?

Approaches

Approach 1: Recursive Depth-First Search (DFS)

Idea:

1. Base Case:
   If the node is null or it does not have children (i.e., it is a leaf), return.

2. Connect Children:
   For any node, connect:

   • node.left.next = node.right
   • If node.next exists, connect node.right.next = node.next.left

3. Recurse:
   Recursively apply the process to the left and right children.

Pros & Cons:

• Pros: Simple recursive solution that directly uses the tree’s structure.
• Cons: Uses recursion, so the recursion depth equals the height of the tree. However, since the tree is perfect, the height is logarithmic.

Approach 2: Iterative Level-by-Level Traversal (Constant Space)

Idea:

1. Start at the Root:
   Use the already established next pointers to traverse each level.
2. Connect Nodes at Next Level:
   For each node at the current level, connect its children using the same rules as in the recursive approach.
3. Move to the Next Level:
   Once a level is processed, move down to the leftmost node of the next level.

Pros & Cons:

• Pros: This approach uses constant extra space (no recursion stack or additional data structures) because it leverages the next pointers for traversal.
• Cons: Slightly more complex to implement than the recursive approach.

Code Implementation

Python Code