103. 二叉树的锯齿形层序遍历--🀄️

categories: [Blog,Algorithm,mid]

103. 二叉树的锯齿形层序遍历

难度中等402
给定一个二叉树，返回其节点值的锯齿形层序遍历。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。
例如：
给定二叉树 [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7

返回锯齿形层序遍历如下：
[
[3],
[20,9],
[15,7]
]

    public List<List<Integer>> zigzagLevelOrder1(TreeNode root) {
        List<List<Integer>> ans = new LinkedList<List<Integer>>();
        if (root == null) {
            return ans;
        }
        Queue<TreeNode> nodeQueue = new LinkedList<TreeNode>();
        nodeQueue.offer(root);
        boolean isOrderLeft = true;
        while (!nodeQueue.isEmpty()) {
            Deque<Integer> levelList = new LinkedList<Integer>();
            int size = nodeQueue.size();
            for (int i = 0; i < size; ++i) {
                TreeNode curNode = nodeQueue.poll();
                if (isOrderLeft) {
                    levelList.offerLast(curNode.val);
                } else {
                    levelList.offerFirst(curNode.val);
                }
                if (curNode.left != null) {
                    nodeQueue.offer(curNode.left);
                }
                if (curNode.right != null) {
                    nodeQueue.offer(curNode.right);
                }
            }
            ans.add(new LinkedList<Integer>(levelList));
            isOrderLeft = !isOrderLeft;
        }
        return ans;
    }

     public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> ans = new LinkedList<List<Integer>>();
        if (root == null) {
            return ans;
        }
        Queue<TreeNode> nodeQueue = new LinkedList<TreeNode>();
        nodeQueue.offer(root);
        boolean isOrderLeft = true;
        while (!nodeQueue.isEmpty()) {
            // Deque<Integer> levelList = new LinkedList<Integer>();
            List<Integer> levelList = new ArrayList<Integer>();
            int size = nodeQueue.size();
            for (int i = 0; i < size; ++i) {
                TreeNode curNode = nodeQueue.poll();
                levelList.add(curNode.val);
                // if (isOrderLeft) {
                //     levelList.offerLast(curNode.val);
                // } else {
                //     levelList.offerFirst(curNode.val);
                // }
                if (curNode.left != null) {
                    nodeQueue.offer(curNode.left);
                }
                if (curNode.right != null) {
                    nodeQueue.offer(curNode.right);
                }
            }
            // ans.add(new ArrayList<Integer>(levelList));
            //isOrderLeft = !isOrderLeft;
            if(ans.size()%2==1){
                Collections.reverse(levelList);
            }
            ans.add(new ArrayList<Integer>(levelList));
        }
        return ans;
    }