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]
]

思路一(dfs)

• 和level order traversal 类似
• 判断
  • 当level % 2 == 0 , 顺序
  • else, 逆序

    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        helper(root, result, 0);
        return result;
    }

    private void helper(TreeNode root, List<List<Integer>> result, int level) {
        if (root == null) return;

        if (level >= result.size()) {
            result.add(new LinkedList<Integer>());
        }
        if (level % 2 == 0) {
            result.get(level).add(root.val);
        } else {
            result.get(level).add(0, root.val);
        }

        helper(root.left, result, level + 1);
        helper(root.right, result, level + 1);
    }

思路二(bfs)

    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        int level = 0;
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> list = new LinkedList<>();
            for (int i = 0; i < size; i++) {
                TreeNode curt = queue.poll();
                if (level % 2 == 0) {
                    list.add(curt.val);
                } else {
                    list.add(0, curt.val);
                }
                if (curt.left != null) {
                    queue.offer(curt.left);
                }
                if (curt.right != null) {
                    queue.offer(curt.right);
                }
            }
            result.add(list);
            level++;
        }
        return result;
    }