Binary Tree Zigzag Level Order Traversal

LeetCode 103 | Difficulty: Medium

Medium

Problem Description

Given the root of a binary tree, return the zigzag level order traversal of its nodes' values. (i.e., from left to right, then right to left for the next level and alternate between).

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [[3],[20,9],[15,7]]

Example 2:

Input: root = [1]
Output: [[1]]

Example 3:

Input: root = []
Output: []

Constraints:

- The number of nodes in the tree is in the range `[0, 2000]`.

- `-100 <= Node.val <= 100`

Topics: Tree, Breadth-First Search, Binary Tree

---

Approach

Tree BFS (Level-Order)

Use a queue to process the tree level by level. At each level, process all nodes in the queue, then add their children. Track the level size to know when one level ends and the next begins.

When to use

Level-order traversal, level-based aggregation, right/left side view.

---

Solutions

Solution 1: C# (Best: 475 ms)

Metric	Value
Runtime	475 ms
Memory	N/A
Date	2017-12-05

Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public IList<IList<int>> ZigzagLevelOrder(TreeNode root) {
    IList<IList<int>> levelOrder = new List<IList<int>>();
    Stack<TreeNode> stack1 = new Stack<TreeNode>();
    Stack<TreeNode> stack2 = new Stack<TreeNode>();
    if (root != null) stack1.Push(root);
    bool isOddRow = true;
    while (stack1.Count != 0 || stack2.Count != 0)
    {
        List<int> levelNums = new List<int>();
        if (isOddRow)
        {
            var levelSize = stack1.Count;
            while(stack1.Count !=0)
            {
                var dequeued = stack1.Pop();
                levelNums.Add(dequeued.val);
                if (dequeued.left != null) stack2.Push(dequeued.left);
                if (dequeued.right != null) stack2.Push(dequeued.right);
            }
        }
        else
        {
            while(stack2.Count != 0)
            {
                var poppedElement = stack2.Pop();
                levelNums.Add(poppedElement.val);
                                if (poppedElement.right != null) stack1.Push(poppedElement.right);
                if (poppedElement.left != null) stack1.Push(poppedElement.left);


            }
        }
        isOddRow  = !isOddRow;
        levelOrder.Add(levelNums);

    }
    return levelOrder;
    }
}

---

Complexity Analysis

Approach	Time	Space
Tree Traversal	$O(n)$	$O(h)$

---

Interview Tips

Key Points

• Discuss the brute force approach first, then optimize. Explain your thought process.
• Consider: "What information do I need from each subtree?" — this defines your recursive return value.