Populating Next Right Pointers in Each Node II LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Populating Next Right Pointers in Each Node II in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

Link for the Problem – Populating Next Right Pointers in Each Node II– LeetCode Problem

Populating Next Right Pointers in Each Node II– LeetCode Problem

Problem:

Given a binary tree

struct Node {
  int val;
  Node *left;
  Node *right;
  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.

Example 1:

117 sample
Input: root = [1,2,3,4,5,null,7]
Output: [1,#,2,3,#,4,5,7,#]
Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.

Example 2:

Input: root = []
Output: []

Constraints:

  • The number of nodes in the tree is in the range [0, 6000].
  • -100 <= Node.val <= 100
Populating Next Right Pointers in Each Node II– LeetCode Solutions
Populating Next Right Pointers in Each Node II Solution in C++:
class Solution {
 public:
  Node* connect(Node* root) {
    Node* node = root;  // the node just above current needling

    while (node) {
      Node dummy(0);  // dummy node before needling
      // needle children of node
      for (Node* needle = &dummy; node; node = node->next) {
        if (node->left) {  // needle left child
          needle->next = node->left;
          needle = needle->next;
        }
        if (node->right) {  // needle right child
          needle->next = node->right;
          needle = needle->next;
        }
      }
      node = dummy.next;  // move node to the next level
    }

    return root;
  }
};
Populating Next Right Pointers in Each Node II Solution in Java:
class Solution {
  public Node connect(Node root) {
    Node node = root; // the node just above current needling

    while (node != null) {
      Node dummy = new Node(); // dummy node before needling
      // needle children of node
      for (Node needle = dummy; node != null; node = node.next) {
        if (node.left != null) { // needle left child
          needle.next = node.left;
          needle = needle.next;
        }
        if (node.right != null) { // needle right child
          needle.next = node.right;
          needle = needle.next;
        }
      }
      node = dummy.next; // move node to the next level
    }

    return root;
  }
}
Populating Next Right Pointers in Each Node II Solution in Python:
class Solution:
  def connect(self, root: 'Node') -> 'Node':
    node = root  # the node just above current needling

    while node:
      dummy = Node(0)  # dummy node before needling
      # needle children of node
      needle = dummy
      while node:
        if node.left:  # needle left child
          needle.next = node.left
          needle = needle.next
        if node.right:  # needle right child
          needle.next = node.right
          needle = needle.next
        node = node.next
      node = dummy.next  # move node to the next level

    return root

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