Solution.java

import javafx.util.Pair;

import java.util.ArrayList;
import java.util.List;

/// Leetcode 102. Binary Tree Level Order Traversal
/// https://leetcode.com/problems/binary-tree-level-order-traversal/description/
/// 二叉树的层序遍历
///
/// 二叉树的层序遍历是一个典型的可以借助队列解决的问题。
/// 该代码主要用于使用Leetcode上的问题测试我们的LinkedListQueue。
/// 对于二叉树的层序遍历，这个课程后续会讲到。
/// 届时，同学们也可以再回头看这个代码。
/// 不过到时，大家应该已经学会自己编写二叉树的层序遍历啦：）

class Solution {

    /// Definition for a binary tree node.
    private class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;
        TreeNode(int x) { val = x; }
    }

    private interface Queue<E> {

        int getSize();
        boolean isEmpty();
        void enqueue(E e);
        E dequeue();
        E getFront();
    }

    private class LinkedListQueue<E> implements Queue<E> {

        private class Node{
            public E e;
            public Node next;

            public Node(E e, Node next){
                this.e = e;
                this.next = next;
            }

            public Node(E e){
                this(e, null);
            }

            public Node(){
                this(null, null);
            }

            @Override
            public String toString(){
                return e.toString();
            }
        }

        private Node head, tail;
        private int size;

        public LinkedListQueue(){
            head = null;
            tail = null;
            size = 0;
        }

        @Override
        public int getSize(){
            return size;
        }

        @Override
        public boolean isEmpty(){
            return size == 0;
        }

        @Override
        public void enqueue(E e){
            if(tail == null){
                tail = new Node(e);
                head = tail;
            }
            else{
                tail.next = new Node(e);
                tail = tail.next;
            }
            size ++;
        }

        @Override
        public E dequeue(){
            if(isEmpty())
                throw new IllegalArgumentException("Cannot dequeue from an empty queue.");

            Node retNode = head;
            head = head.next;
            retNode.next = null;
            if(head == null)
                tail = null;
            size --;
            return retNode.e;
        }

        @Override
        public E getFront(){
            if(isEmpty())
                throw new IllegalArgumentException("Queue is empty.");
            return head.e;
        }

        @Override
        public String toString(){
            StringBuilder res = new StringBuilder();
            res.append("Queue: front ");

            Node cur = head;
            while(cur != null)
                res.append(cur + "->");
            res.append("NULL tail");
            return res.toString();
        }
    }

    public List<List<Integer>> levelOrder(TreeNode root) {

        ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();
        if(root == null)
            return res;

        // 我们使用LinkedList来做为我们的先入先出的队列
        LinkedListQueue<Pair<TreeNode, Integer>> queue = new LinkedListQueue<Pair<TreeNode, Integer>>();
        queue.enqueue(new Pair<TreeNode, Integer>(root, 0));

        while(!queue.isEmpty()){

            Pair<TreeNode, Integer> front = queue.dequeue();
            TreeNode node = front.getKey();
            int level = front.getValue();

            if(level == res.size())
                res.add(new ArrayList<Integer>());
            assert level < res.size();

            res.get(level).add(node.val);
            if(node.left != null)
                queue.enqueue(new Pair<TreeNode, Integer>(node.left, level + 1));
            if(node.right != null)
                queue.enqueue(new Pair<TreeNode, Integer>(node.right, level + 1));
        }

        return res;
    }
}