103. Binary Tree Zigzag Level Order Traversal

Solution 1：

public class Solution {

    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {

        List<List<Integer>> sol = new ArrayList<>();

        travel(root, sol, 0);

        return sol;

    }

    

    private void travel(TreeNode curr, List<List<Integer>> sol, int level) {

        if (curr == null)

            return;

        if (sol.size() <= level) {            

            sol.add(new ArrayList<>());

        }

        

        if (level % 2 == 0)

            sol.get(level).add(curr.val);

        else 

            sol.get(level).add(0, curr.val); //在0位置直接add，所以原来的元素右移

        

        travel(curr.left, sol, level + 1);

        travel(curr.right, sol, level + 1);

    }

}

Solution 2：

public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
    List<List<Integer>> res = new ArrayList<>();
    if (root == null)
        return res;
    Queue<TreeNode> q = new LinkedList<>();
    q.add(root);
    boolean order = true;

    while (!q.isEmpty()) {
        int size = q.size();
        List<Integer> tmp = new ArrayList<>();
        for (int i = 0; i < size; ++i) {
            TreeNode n = q.poll();
            if (order) {
                tmp.add(n.val);
            } else {
                tmp.add(0, n.val);
            }
            if (n.left != null) q.add(n.left);
            if (n.right != null) q.add(n.right);
        }
        res.add(tmp);
        order = !order;
    }
    return res;
}