Binary Tree Zigzag Level Order Traversal

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https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/

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]
]

Solution:

  • Level-traversal using a Queue + DFS
  • On enqueue put the elements on a list
  • Apply list.reverse for the odd-levels
Works fast. Code's below, cheers, ACC.



public class Solution
{
    public IList<IList<int>> ZigzagLevelOrder(TreeNode root)
    {
        if (root == null) return new List<IList<int>>();

        List<IList<int>> retVal = new List<IList<int>>();

        Queue<QueueItem> queue = new Queue<QueueItem>();
        QueueItem qi = new QueueItem(root, 0);
        int currentLevel = 0;
        queue.Enqueue(qi);

        List<int> currentList = new List<int>();
        int count = 0;
        while (queue.Count > 0)
        {
            QueueItem currentItem = queue.Dequeue();

            if (currentLevel < currentItem.level)
            {
                if (count % 2 == 1)
                {
                    currentList.Reverse();
                }
                retVal.Add(currentList);
                currentList = new List<int>();
                count++;
            }
            currentLevel = currentItem.level;
            currentList.Add(currentItem.node.val);

            if (currentItem.node.left != null)
            {
                QueueItem nextItem = new QueueItem(currentItem.node.left, currentItem.level + 1);
                queue.Enqueue(nextItem);
            }
            if (currentItem.node.right != null)
            {
                QueueItem nextItem = new QueueItem(currentItem.node.right, currentItem.level + 1);
                queue.Enqueue(nextItem);
            }
        }
        if (currentList.Count > 0)
        {
            if (count % 2 == 1)
            {
                currentList.Reverse();
            }
            retVal.Add(currentList);
        }

        return retVal;
    }
}

public class QueueItem
{
    public TreeNode node = null;
    public int level = 0;

    public QueueItem(TreeNode node, int level)
    {
        this.node = node;
        this.level = level;
    }
}

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