116. Populating Next Right Pointers in Each Node.cpp

/*
Link: https://leetcode.com/problems/populating-next-right-pointers-in-each-node/submissions/

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

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.

 

Follow up:

You may only use constant extra space.
Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.
 

Example 1:



Input: root = [1,2,3,4,5,6,7]
Output: [1,#,2,3,#,4,5,6,7,#]
Explanation: Given the above perfect 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.
 

Constraints:

The number of nodes in the given tree is less than 4096.
-1000 <= node.val <= 1000

*/

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() : val(0), left(NULL), right(NULL), next(NULL) {}

    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

    Node(int _val, Node* _left, Node* _right, Node* _next)
        : val(_val), left(_left), right(_right), next(_next) {}
};
*/
// Solution 1 : Iterative using queue
/*
TC = O(n)
SC = O(n)
*/
class Solution {
public:
    Node* connect(Node* root) {
        if (!root) {
            return root;
        }
        
        queue<Node *> q;                                // queue for storing the nodes
                
        q.push(root);                                   // pushing node
        
        
        Node * temp = NULL;
        
        while (!q.empty()) {                            
            int len = q.size();                         // number of nodes at previous level
            
            for (int i = 0; i < len; i++) {             // running the loop q.size() times, which is the number of nodes in the previous level
                
                Node *prev = temp;                      // The left node to the current node
                
                temp = q.front();
                q.pop();
                
                if (i > 0) {                            // ignoring when the node it the leftmost node
                    prev -> next = temp;
                }
                
                if (temp -> left) {                     // if node has left child
                    q.push(temp -> left);
                }
                
                if (temp -> right) {
                    q.push(temp -> right);
                }
            }
        }
        return root;
    }
};