[leedcode 103] 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 between).
For example:
Given binary tree {3,9,20,#,#,15,7}
,
3 / \ 9 20 / \ 15 7
return its zigzag level order traversal as:
[ [3], [20,9], [15,7] ]
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<List<Integer>> zigzagLevelOrder(TreeNode root) { //和正常的层序遍历的唯一不同是,增加一个标志flag,代表从左还是从右开始遍历,使用了Collections的静态方法reverse反转链表 List<List<Integer>> res=new ArrayList<List<Integer>>(); List<Integer> seq=new ArrayList<Integer>(); if(root==null) return res; LinkedList<TreeNode> queue=new LinkedList<TreeNode>(); queue.add(root); int cutcont=1; int nextcont=0; boolean flag=false; while(!queue.isEmpty()){ TreeNode node=queue.remove(); seq.add(node.val); cutcont--; if(node.left!=null){ queue.add(node.left); nextcont++; } if(node.right!=null){ queue.add(node.right); nextcont++; } /* while(cutcont>0){ TreeNode node=queue.remove(); seq.add(node.val); cutcont--; if(node.left!=null){ queue.add(node.left); nextcont++; } if(node.right!=null){ queue.add(node.right); nextcont++; } }*/ if(cutcont==0){ if(flag){ Collections.reverse(seq); } flag=!flag; res.add(seq); cutcont=nextcont; nextcont=0; seq=new ArrayList<Integer>(); } } return res; } }