问题描述
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,¥,¥,15,7},
3
/
9 20
/
15 7return 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
/
45
The above binary tree is serialized as”{1,2,3,¥,¥,4,¥,¥,5}”.
问题分析
这题主要还是考层序遍历,但是每层可能正序输出和倒序输出,所以我们最好用栈或者最后倒序一下输出
代码实现
public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
ArrayList<ArrayList<Integer>> result = new ArrayList<>();
if (root == null) return result;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (queue.isEmpty() == false) {
ArrayList<Integer> list = new ArrayList<>();
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode curNode = queue.poll();
list.add(curNode.val);
if (curNode.left != null) queue.offer(curNode.left);
if (curNode.right != null) queue.offer(curNode.right);
}
result.add(list);
}
//处理锯齿形的输出
ArrayList<ArrayList<Integer>> res = new ArrayList<>();
int count = 0;
while (count < result.size()) {
//一行正输出,一行反输出
ArrayList<Integer> list = new ArrayList<>();
if (count % 2 == 0) {
for (int i = 0; i < result.get(count).size(); i++) {
list.add(result.get(count).get(i));
}
} else if (count % 2 == 1) {
for (int i = 0; i < result.get(count).size(); i++) {
list.add(0, result.get(count).get(i));
}
}
count++;
res.add(list);
}
return res;
}