LeetCode Binary Tree Zigzag Level Order Traversal

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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]]

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 '#' signifies a path terminator where no node exists below.

Here's an example:

   1  / \ 2   3    /   4    \     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

题意：层次遍历一颗树，奇数层的时候翻转。

思路：还是利用队列层次遍历，多一个判断是否翻转的标记就行了，还有就是java是引用传递的，所以每次都要重新声明一个对象，不然ans里面都是指向同一块内存。

/** * Definition for binary tree * 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>> ans = new ArrayList<List<Integer>>();    if (root == null) return ans;        Queue<TreeNode> queue = new LinkedList<TreeNode>();    queue.add(root);    boolean reverse = false;    while (!queue.isEmpty()) {    List<Integer> tmp = new ArrayList<Integer>();    int num = queue.size();    for (int i = 0; i < num; i++) {    TreeNode t = queue.poll();    tmp.add(t.val);    if (t.left != null)     queue.add(t.left);    if (t.right != null)     queue.add(t.right);    }    if (reverse) {    Collections.reverse(tmp);    reverse = false;    } else reverse = true;    ans.add(tmp);    }        return ans;     }}

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