Binary Tree Zigzag Level Order Traversal @ LeetCode

It is also a variant of Level Order Traversal. It can be seen that it is necessary to write Level Order Traversal skillfully!

Package Level4; import java. util. arrayList; import java. util. collections; import java. util. using list; import java. util. queue; import Utility. treeNode;/*** Binary Tree Zigzag Level Order Traversal ** Given a binary tree, return the zigzag level order traversal of its nodes 'values. (ie, from left to right, then right to left for the next level and alternate ). for example: Given binary tr Ee {3, 9, 20, #, #, 15, 7}, 3/\ 9 20/\ 15 7 return its zigzag level order traversal as: [[3], [20, 9], [15, 7] confused what "{1, #, 2, 3}" means?> Read more on how binary tree is serialized on OJ. OJ's Binary Tree Serialization: The serialization of a binary tree follows a level order traversal, where '# 'signiies a path terminator where no node exists below. here's an example: 1/\ 2 3/4 \ 5 The above binary tree is serialized as "{, 3, #,#, 4, #,#, 5 }". */public class S103 {public static void main (String [] args) {} public ArrayList <ArrayList <Integer> zigzagLevelOrder (TreeNode root) {ArrayList <Integer> ret = new ArrayList <Integer> (); if (root = null) {return ret;} Queue <TreeNode> queue = new queue list <TreeNode> (); ArrayList <Integer> al = new ArrayList <Integer> (); Queue. add (root); int currentLevel = 1; int nextLevel = 0; boolean left = true; while (! Queue. isEmpty () {TreeNode cur = queue. remove (); currentLevel --; al. add (cur. val); if (cur. left! = Null) {queue. add (cur. left); nextLevel ++;} if (cur. right! = Null) {queue. add (cur. right); nextLevel ++;} if (currentLevel = 0) {if (! Left) {// when right to left, you need to flip al Collections. reverse (al);} left =! Left; ret. add (al); al = new ArrayList <Integer> (); currentLevel = nextLevel; nextLevel = 0 ;}} return ret ;}}