[LeetCode] 116. Populating Next Right Pointers in Each Node

Problem : https://leetcode.com/problems/populating-next-right-pointers-in-each-node/

BFS solution :

"""
# Definition for a Node.
class Node:
    def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""

class Solution:
    def connect(self, root: 'Node') -> 'Node':
        if not root:
            return None
        
        
        queue = deque([root])
        
        while queue:
            level = deque()
            
            while queue:
                node = queue.popleft()
                
                if node.left:
                    level.append(node.left)
                
                if node.right:
                    level.append(node.right)
            
            for i in range(1, len(level)):
                level[i-1].next = level[i]
                
            queue = level
        
        
        return root

Recursive solution:

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

    public Node() {}
    
    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, Node _left, Node _right, Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/

class Solution {
    public Node connect(Node root) {
        if (root != null) {
            if (root.left != null) {
                root.left.next = root.right;
            }
        
            if (root.right != null && root.next != null) {
                root.right.next = root.next.left;
            }
        
            connect(root.right); 
            connect(root.left);
        }
        
        return root;
    }
}

Iterative solution:

"""
# Definition for a Node.
class Node:
    def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""

class Solution:
    def connect(self, root: 'Node') -> 'Node':
        tmp = root
        while root and root.left:
            nxt = root.left
            
            while root:
                root.left.next = root.right
                root.right.next = root.next.left if root.next else None
                root = root.next
            root = nxt
        
        return tmp

Edited on 12/29/2021. Replace the recursive solution with Java code solution.