package week02;

import java.util.LinkedList;

import java.util.List;

import java.util.Queue;

/**

* @创建人 luoxiang

* @创建时间 2019/6/14 16:24

* @描述 LeetCode : 102. 二叉树的层次遍历 https://leetcode-cn.com/problems/binary-tree-level-order-traversal/

*/

public class BinaryTreeLevelorderTraversal102 {

public static void main(String[] args) {

TreeNode node = new TreeNode(3);

node.left = new TreeNode(5);

node.right = new TreeNode(1);

node.left.left = new TreeNode(6);

node.left.right = new TreeNode(2);

node.left.right.left = new TreeNode(7);

node.left.right.right = new TreeNode(4);

node.right.left = new TreeNode(0);

node.right.right = new TreeNode(8);

final List<List<Integer>> list = new BinaryTreeLevelorderTraversal102().levelOrder(node);

for (List<Integer> integers : list) {

for (Integer integer : integers) {

System.out.print(integer + ",");

}

System.out.println();

}

}

/**

* Method 1 : BFS ，队列存储每一层的值, 用 currentSize 记录当前层的数量，currentSize 为0时，当前层遍历完毕

* 时间复杂度 : O(N) 树的遍历 ; 空间复杂度 ： O(N) ; 额外的空间 队列

*/

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

List<List<Integer>> list = new LinkedList<>();

if (root == null) return list;

Queue<TreeNode> queue = new LinkedList<>();

queue.offer(root);

int currentSize = 1;

List<Integer> currentList = new LinkedList<>();

while (!queue.isEmpty()) {

TreeNode node = queue.poll();

currentList.add(node.val);

if (node.left != null) queue.offer(node.left);

if (node.right != null) queue.offer(node.right);

if (--currentSize == 0) {

currentSize = queue.size();

list.add(currentList);

currentList = new LinkedList<>();

}

}

return list;

}

}