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],

return its zigzag level order traversal as:

[
  [3],
  [20,9],
  [15,7]
]

Solution (Java)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        Queue<TreeNode> q = new LinkedList<>();
        q.add(root);
        q.add(null);
        boolean zig = true;
        LinkedList<Integer> level = new LinkedList<>();
        while (!q.isEmpty()) {
            TreeNode node = q.remove();
            while (!q.isEmpty() && node != null) {
                if (zig) {
                    level.add(node.val);
                } else {
                    level.addFirst(node.val);
                }
                if (node.left != null) {
                    q.add(node.left);
                }
                if (node.right != null) {
                    q.add(node.right);
                }
                node = q.remove();
            }
            result.add(level);
            zig = !zig;
            level = new LinkedList<>();
            if (!q.isEmpty()) {
                q.add(null);
            }
        }
        return result;
    }
}

Solution (Javascript)

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var zigzagLevelOrder = function(root) {
  if (!root) return [];
  return helper([[root]], 0);
};

var helper = function (res, level) {
  var now = res[level];
  var next = [];

  for (var i = now.length - 1; i >= 0; i--) {
    if (level % 2) {
      if (now[i].left) next.push(now[i].left);
      if (now[i].right) next.push(now[i].right);
    } else {
      if (now[i].right) next.push(now[i].right);
      if (now[i].left) next.push(now[i].left);
    }

    now[i] = now[i].val;
  }

  if (next.length) {
    res.push(next);
    helper(res, level + 1);
  }

  return res;
};

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).