116. Populating Next Right Pointers in Each Node

Solution1: O(N) O(N)

Level order traversal

class Solution {
    public Node connect(Node root) {
        if (root == null) {
            return root;
        }
        Queue<Node> queue = new LinkedList<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                Node node = queue.poll();

                if (i != size - 1) {
                    node.next = queue.peek();
                }

                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
        }
        return root;
    }
}

Solution2 : O(N) O(1)

• when we go over the nodes of a particular level, their next pointers are already established.
• This is what helps get rid of the queue data structure from the previous approach and helps save space.
• There are two types of next connextions.
  • Under the same parent.
  • Between different parents

class Solution {
    public Node connect(Node root) {
        if (root == null) {
            return root;
        }

        // first Node at current level
        Node start = root; 

        // loop over each level 
        while (start.left != null) {
            Node cur = start; 
            // traverse the cur level linkedlist, and connect the next level nodes
            while (cur != null) {
                // type1, two next level nodes share same parent
                cur.left.next = cur.right;
                // type2, two next level nodes have different parent
                if (cur.next != null) {
                    cur.right.next = cur.next.left;
                }
                cur = cur.next;
            }
            // move to next level
            start = start.left;
        }
        return root;
    }
}