116. Populating Next Right Pointers in Each Node

Problem

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

Solution

This problem requires us to point the next of the left node in each layer to the right node. Because it is processed layer by layer, it is very intuitive to think of using BFS to solve it. We need two queues so that we can distinguish between the current layer and the next layer. Because the problem mentions that it is a perfect binary tree, the left child of each node can directly point to the right child. Before adding the left child to the queue, point the last node in the queue to the left child.

BFS

• Time: 
• Space: 

class Solution {
    public Node connect(Node root) {
        if (root == null) return null;
        Deque<Node> currentLevel = new ArrayDeque<>();
        Deque<Node> nextLevel = new ArrayDeque<>();
        currentLevel.offer(root);
        
        while (!currentLevel.isEmpty()) {
            Node node = currentLevel.poll();
            if (node.left != null) {
                if (!nextLevel.isEmpty()) {
                    nextLevel.peekLast().next = node.left;
                }
                nextLevel.offer(node.left);
                node.left.next = node.right;
                nextLevel.offer(node.right);
            }

            if (currentLevel.isEmpty()) {
                currentLevel = nextLevel;
                nextLevel = new ArrayDeque<>();
            }
        }
        return root;
    }
}

Reference

• 116. Populating Next Right Pointers in Each Node, LeetCode.
• 116. Populating Next Right Pointers in Each Node, LeetCode Solutions.