Leetcode: Binary Tree Zigzag Level Order Traversal

URL

https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/description/

描述

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,null,null,15,7],

3

/ \
9 20
/ \
15 7

return its zigzag level order traversal as:

[
[3],
[20,9],
[15,7]
]

解题思路

本题使用深度优先搜索或者广度优先搜索，在不同的层次注意顺序即可，下面给出代码。
DFS

/** * 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){     List<List<Integer>> res = new ArrayList<>();        dfs(root,0,res);        return res;    }    public void dfs(TreeNode root,int level,List<List<Integer>> res){        if(root==null) return;        if(res.size()<=level){            res.add(new ArrayList<>());        }        List<Integer> row = res.get(level);        if(level%2==0){            row.add(root.val);        }else{            row.add(0,root.val);        }        dfs(root.left,level+1,res);        dfs(root.right,level+1,res);    }}

BFS

/** * 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) {        List<List<Integer>> res = new ArrayList<>();        if(root==null) return res;        Queue<TreeNode> queue = new LinkedList<TreeNode>();        queue.add(root);        int size = 0;        int counter = 0;        while(!queue.isEmpty()){            size = queue.size();            LinkedList<Integer> row = new LinkedList<>();            for(int i=0;i<size;i++){                TreeNode tmp = queue.poll();                if(counter%2==0){                    row.add(tmp.val);                }else{                    row.addFirst(tmp.val);                }                if(tmp.left!=null)                    queue.add(tmp.left);                if(tmp.right!=null)                    queue.add(tmp.right);            }            counter++;            res.add(row);        }        return res;    }}