L103. Binary Tree Zigzag Level Order Traversal

Problem:

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]
]

Solution:

• set a flag to identify the level num is odd or even;

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