116. Populating Next Right Pointers in Each Node

Problem Statement

116. Populating Next Right Pointers in Each Node

Medium

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.

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.

Example 2:

Input: root = []
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 2¹² - 1].
-1000 <= Node.val <= 1000

Follow-up:

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

C++

Source file

/*
// 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) {}
};
*/

class Solution {
public:
    vector<queue <Node*>> v;    // Vector of Queues
    bool firstTime = true;  // Meme reference : https://knowyourmeme.com/memes/james-franco-first-time
    Node* connect(Node* root, int lvl=0) {
        // Execute in first run only
        if (firstTime){
            if (!(root))
                return root;
            // Constructing empty queues for every level in tree
            auto ptr = root;
            while(ptr->left){
                v.push_back(queue<Node*>());
                ptr = ptr->left;
            }
            firstTime = false;
        }
        auto rootsLeft = root->left;
        auto rootsRight = root->right;
        if (rootsLeft){
            rootsLeft->next = rootsRight;
            if (!(v[lvl].empty())){// leftmost nodes are not the next of any node
            (v[lvl].front())->next = rootsLeft;
            v[lvl].pop();
            }
            connect(rootsLeft, lvl+1);
        }
        if (rootsRight){
            // next of tree's right-most border remains null
            v[lvl].push(rootsRight);
            connect(rootsRight, lvl+1);
        }
        return root;
    }
};