103. Binary Tree Zigzag Level Order Traversal

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.

---

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {

ArrayList<ArrayList<Integer>> rst = new ArrayList<ArrayList<Integer>>();
ArrayList<Integer> curList = new ArrayList<Integer>();
ArrayList<TreeNode> curNodeList = new ArrayList<TreeNode>();

// check
if(root==null)  return rst;

// init root
curList.add(root.val);
curNodeList.add(root);
int level = 0;

while(!curNodeList.isEmpty()){

// add previous level to rst, reverse in odd level
if(level %2 != 0)
Collections.reverse(curList);
rst.add(curList);

// update
ArrayList<TreeNode> preNodeList = curNodeList;
curNodeList = new ArrayList<TreeNode>();
curList = new ArrayList<Integer>();

// go through the prevois level
for(TreeNode n : preNodeList){
if(n.left != null){
curNodeList.add(n.left);
curList.add(n.left.val);
}
if(n.right != null){
curNodeList.add(n.right);
curList.add(n.right.val);
}
}
level++;
}

return rst;

}
}