package week02;

import java.util.Collections;

import java.util.LinkedList;

import java.util.List;

import java.util.Queue;

/**

* @创建人 luoxiang

* @创建时间 2019/6/14 16:41

* @描述 LeetCode : 103. 二叉树的锯齿形层次遍历 https://leetcode-cn.com/problems/binary-tree-zigzag-level-order-traversal/

*/

public class BinaryTreeZigzagLevelOrderTraversal103 {

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 BinaryTreeZigzagLevelOrderTraversal103().zigzagLevelOrder(node);

for (List<Integer> integers : list) {

for (Integer integer : integers) {

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

}

System.out.println();

}

}

/**

* Method 1 : 对 准备新插入 List<Integer> 进行处理，用一个奇偶数 或 布尔值 来记录是否需要反转

* 时间复杂度 ： O(N) 树的遍历

* 空间复杂度 : O(N) 额外空间 队列

*

*/

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

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

if (root == null) return list;

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

queue.offer(root);

int currentSize = 1;

int depth =0;

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();

if(depth%2!=0) Collections.reverse(currentList);

list.add(currentList);

currentList = new LinkedList<>();

depth++;

}

}

return list;

}

public List<List<Integer>> zigzagLevelOrder2(TreeNode root) {

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

if (root == null) return list;

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

queue.offer(root);

int currentSize = 1;

boolean depth =false;

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();

if(depth) Collections.reverse(currentList);

list.add(currentList);

currentList = new LinkedList<>();

depth=!depth;

}

}

return list;

}

}