[LeetCode][Java] Binary Tree Zigzag Level Order Traversal

题目：

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]
]

题意：

给定一棵二叉树，返回这棵树的节点的'Z'字形遍历（即 起始顺序为先左后右，下一层的顺序为先右后左，这样交替进行）

比如

给定二叉树

{3,9,20,#,#,15,7}

,

3
/ \
9  20
/  \
15   7

返回这棵树的'z'字形遍历：

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

算法分析：

利用题目《Binary Tree Level Order Traversal》的结果，对ArrayList中的奇数元素进行倒序 。难度不大，直接上代码

AC代码：

<span style="font-family:Microsoft YaHei;font-size:12px;">public class Solution
{
public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root)
{
ArrayList<ArrayList<Integer>> list = new ArrayList<ArrayList<Integer>>();
if (root == null)
return list;

list = levelOrder(root);
for(int i=1;i<list.size();i+=2)
{
reverse(list.get(i));
}
return list;
}
private static void reverse(ArrayList<Integer> temlist)
{
int tem=0;
int temsize=0;
if(temlist.size()%2==0)
temsize=temlist.size()/2-1;
else
temsize=temlist.size()/2;
for(int i=0;i<=temsize;i++)
{
tem=temlist.get(i);
temlist.set(i, temlist.get(temlist.size()-1-i));
temlist.set(temlist.size()-1-i, tem);
}

}
public static ArrayList<ArrayList<Integer>> levelOrder(TreeNode root)
{
ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
if (root == null)
{
return res;
}
ArrayList<Integer> tmp = new ArrayList<Integer>();
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
int num;
boolean reverse = false;
while (!queue.isEmpty())
{
num = queue.size();  //每次通过这个确定最终的出队数目
tmp.clear();
for (int i = 0; i < num; i++) //队列中出1个父，进两个子；出2个父，进4个子；出4个父，进8个子
{
TreeNode node = queue.poll();
tmp.add(node.val);
if (node.left != null)
queue.offer(node.left);
if (node.right != null)
queue.offer(node.right);
}

res.add(new ArrayList<Integer>(tmp));
}
return res;
}
}</span>