103. Binary Tree Zigzag Level Order Traversal

2020-01-27

Recursive

We don’t traverse the tree in zigzag level order but traverse in normal level order. We just reverse the list which has odd height index then return the result.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {

    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        LinkedList<List<Integer>> res = new LinkedList<>();
        if (root == null) {
            return res;
        }
        helper(root, 0, res);
        for (int i = 0; i < res.size(); i++) {
            if (i % 2 != 0) {
                Collections.reverse(res.get(i));
            }
        }
        return res;
    }

    private void helper(TreeNode root, int height, List<List<Integer>> res) {
        if (root == null) {
            return;
        }
        if (res.size() <= height) {
            res.add(new ArrayList<>());
        }
        res.get(height).add(root.val);
        helper(root.left, height + 1, res);
        helper(root.right, height + 1, res);
    }
}

Iterative

In the iterative solution, we still traverse the tree in normal level order, but use a different way to get the result.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        LinkedList<List<Integer>> res = new LinkedList<>();
        if (root == null) {
            return res;
        }

        LinkedList<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        int height = 0;
        while (!queue.isEmpty()) {
            // add a empty list for the current level
            res.add(new ArrayList<>());

            // add all elements to the list
            ArrayDeque<Integer> deque = new ArrayDeque<>();
            int elementLen = queue.size();
            for (int i = 0; i < elementLen; i++) {
                TreeNode node = queue.removeFirst();
                if (height % 2 == 0) {
                    deque.addLast(node.val);
                } else {
                    deque.addFirst(node.val);
                }
                if (node.left != null) {
                    queue.addLast(node.left);
                }
                if (node.right != null) {
                    queue.addLast(node.right);
                }
            }
            res.get(height).addAll(deque);

            // go to the next level
            height++;
        }

        return res;
    }
}