117. Populating Next Right Pointers in Each Node II

Problem

Given a binary tree

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,null,7]
Output: [1,#,2,3,#,4,5,7,#]
Explanation: Given the above 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, 6000].
• -100 <= Node.val <= 100

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. Before adding each child to the queue, point the last node in the queue to the 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);
            }
            if (node.right != null) {
                if (!nextLevel.isEmpty()) {
                    nextLevel.peekLast().next = node.right;
                }
                nextLevel.offer(node.right);
            }

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

Reference

• 117. Populating Next Right Pointers in Each Node II, LeetCode.
• 117. Populating Next Right Pointers in Each Node II, LeetCode Solution.