103. Binary Tree Zigzag Level Order Traversal

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).

Examples

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: []
Jump to Solution: Python Java C++ JavaScript C#

Python Solution


# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
from collections import deque

def zigzagLevelOrder(root: Optional[TreeNode]) -> List[List[int]]:
    if not root:
        return []
    
    result = []
    queue = deque([root])
    left_to_right = True
    
    while queue:
        level_size = len(queue)
        current_level = []
        
        for _ in range(level_size):
            node = queue.popleft()
            
            # Add to current level based on direction
            if left_to_right:
                current_level.append(node.val)
            else:
                current_level.insert(0, node.val)
            
            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)
        
        result.append(current_level)
        left_to_right = not left_to_right
    
    return result

Java Solution


/**
 * 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> zigzagLevelOrder(TreeNode root) {
        List> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        
        Queue queue = new LinkedList<>();
        queue.offer(root);
        boolean leftToRight = true;
        
        while (!queue.isEmpty()) {
            int levelSize = queue.size();
            LinkedList currentLevel = new LinkedList<>();
            
            for (int i = 0; i < levelSize; i++) {
                TreeNode node = queue.poll();
                
                // Add to current level based on direction
                if (leftToRight) {
                    currentLevel.add(node.val);
                } else {
                    currentLevel.addFirst(node.val);
                }
                
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
            
            result.add(currentLevel);
            leftToRight = !leftToRight;
        }
        
        return result;
    }
}

C++ Solution


/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector> zigzagLevelOrder(TreeNode* root) {
        vector> result;
        if (!root) {
            return result;
        }
        
        queue queue;
        queue.push(root);
        bool leftToRight = true;
        
        while (!queue.empty()) {
            int levelSize = queue.size();
            vector currentLevel(levelSize);
            
            for (int i = 0; i < levelSize; i++) {
                TreeNode* node = queue.front();
                queue.pop();
                
                // Add to current level based on direction
                int index = leftToRight ? i : levelSize - 1 - i;
                currentLevel[index] = node->val;
                
                if (node->left) {
                    queue.push(node->left);
                }
                if (node->right) {
                    queue.push(node->right);
                }
            }
            
            result.push_back(currentLevel);
            leftToRight = !leftToRight;
        }
        
        return result;
    }
};

JavaScript Solution


/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var zigzagLevelOrder = function(root) {
    if (!root) {
        return [];
    }
    
    const result = [];
    const queue = [root];
    let leftToRight = true;
    
    while (queue.length) {
        const levelSize = queue.length;
        const currentLevel = [];
        
        for (let i = 0; i < levelSize; i++) {
            const node = queue.shift();
            
            // Add to current level based on direction
            if (leftToRight) {
                currentLevel.push(node.val);
            } else {
                currentLevel.unshift(node.val);
            }
            
            if (node.left) {
                queue.push(node.left);
            }
            if (node.right) {
                queue.push(node.right);
            }
        }
        
        result.push(currentLevel);
        leftToRight = !leftToRight;
    }
    
    return result;
};

C# Solution


/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public IList> ZigzagLevelOrder(TreeNode root) {
        IList> result = new List>();
        if (root == null) {
            return result;
        }
        
        Queue queue = new Queue();
        queue.Enqueue(root);
        bool leftToRight = true;
        
        while (queue.Count > 0) {
            int levelSize = queue.Count;
            LinkedList currentLevel = new LinkedList();
            
            for (int i = 0; i < levelSize; i++) {
                TreeNode node = queue.Dequeue();
                
                // Add to current level based on direction
                if (leftToRight) {
                    currentLevel.AddLast(node.val);
                } else {
                    currentLevel.AddFirst(node.val);
                }
                
                if (node.left != null) {
                    queue.Enqueue(node.left);
                }
                if (node.right != null) {
                    queue.Enqueue(node.right);
                }
            }
            
            result.Add(currentLevel.ToList());
            leftToRight = !leftToRight;
        }
        
        return result;
    }
}

Complexity Analysis

Solution Explanation

This solution uses breadth-first search with alternating direction:

Key points: