leetcode第46题（binary-tree-zigzag-level-order-traversal）

题目：

Given a binary tree, return the zigzag level ordertraversal 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}". 

思路：用两个队列或者两个堆实现。或者是用一个队列或者堆，奇数行的时候实现翻转即可。

代码：

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
import java.util.ArrayList;
import java.util.Collections;
import java.util.LinkedList;
import java.util.Queue;
public class Solution {
public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
ArrayList<ArrayList<Integer>> result = new ArrayList<>();
Queue<TreeNode&
9a02
gt; queue = new LinkedList<>();
boolean left = true;
if (root != null) {
queue.add(root);
}
while (!queue.isEmpty()) {
int n = queue.size();
ArrayList<Integer> temp = new ArrayList<>();
for(int i=n-1;i>=0;i--) {
TreeNode node = queue.poll();
if (node.left != null) {
queue.add(node.left);
}
if (node.right != null) {
queue.add(node.right);
}
temp.add(node.val);
}
if (result.size() % 2 != 0) {
Collections.reverse(temp);
}
result.add(temp);
}
return result;
}
}