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,null,null,15,7]

,

3
/ \
9  20
/  \
15   7

return its zigzag level order traversal as:

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

/**
* Definition for a binary tree node.
* 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>> res = new ArrayList<>();
if(root == null)
return res;
Queue<TreeNode> treeQueue = new ArrayDeque<>();
treeQueue.offer(root);
Stack<Integer> stack = new Stack<>();
Queue<Integer> level = new ArrayDeque<>();
Queue<Integer> queue = new ArrayDeque<>();
int depth = 1;
level.offer(depth);
int lastLevel = 1;
while(!treeQueue.isEmpty()){
TreeNode current = treeQueue.poll();
int currentLevel = level.poll();
if(currentLevel != lastLevel){
List<Integer> list = new ArrayList<>();
if(lastLevel % 2 == 0){
while(!stack.isEmpty()){
list.add(stack.pop());
}
res.add(list);
}else{
while(!queue.isEmpty()){
list.add(queue.poll());
}
res.add(list);
}
}

if(currentLevel % 2 == 0){
stack.push(current.val);
}else{
queue.offer(current.val);
}

lastLevel = currentLevel;
if(current.left != null){
treeQueue.offer(current.left);
level.offer(currentLevel+1);
}
if(current.right != null){
treeQueue.offer(current.right);
level.offer(currentLevel+1);
}
}
List<Integer> list = new ArrayList<>();
if(lastLevel % 2 == 0){
while(!stack.isEmpty()){
list.add(stack.pop());
}
res.add(list);
}else{
while(!queue.isEmpty()){
list.add(queue.poll());
}
res.add(list);
}
return res;
}
}