[LC] 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,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.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def zigzagLevelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        from collections import deque
        res = []
        if root is None:
            return res
        is_even = False
        my_deque = deque()
        my_deque.append(root)
        while my_deque:
            size = len(my_deque)
            lst = []
            for _ in range(size):
                if is_even:
                    cur = my_deque.pop()
                    if cur.right:
                        my_deque.appendleft(cur.right)
                    if cur.left:
                        my_deque.appendleft(cur.left)
                else:
                    cur = my_deque.popleft()
                    if cur.left:
                        my_deque.append(cur.left)
                    if cur.right:
                        my_deque.append(cur.right)
                lst.append(cur.val)
            res.append(lst)
            is_even = not is_even
        return res

 

    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        
        boolean isReverse = false;
        Deque<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()) {
            int len = queue.size();
            List<Integer> list = new ArrayList<>();
            for (int i = 0; i < len; i++) {
                if (isReverse) {
                    TreeNode cur = queue.pollFirst();
                    list.add(cur.val);
                    if (cur.right != null) {
                        queue.offerLast(cur.right);
                    }
                    if (cur.left != null) {
                        queue.offerLast(cur.left);
                    }
                } else {    
                    TreeNode cur = queue.pollLast();
                    list.add(cur.val);
                    if (cur.left != null) {
                        queue.offerFirst(cur.left);
                    }
                    if (cur.right != null) {
                        queue.offerFirst(cur.right);
                    }
                }   
            }
            isReverse = !isReverse;
            res.add(list);
        }
        return res;
    }