---
tags: leetcode
---

# [103. Binary Tree Zigzag Level Order Traversal](https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/)
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# My Solution
## The Key Idea for Solving This Coding Question
BFS
## C++ Code

/**
 * 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<vector<int>> zigzagLevelOrder(TreeNode *root) {
        if (root == nullptr) {
            return {};
        }

        queue<TreeNode *> q;
        vector<vector<int>> zigzag;
        bool reverse = false;

        q.push(root);
        while (!q.empty()) {
            int qLen = q.size();
            vector<int> oneLevel(qLen);

            for (int i = 0; i < qLen; ++i) {
                TreeNode *curr = q.front();
                q.pop();

                if (reverse) {
                    oneLevel[oneLevel.size() - 1 - i] = curr->val;
                } else {
                    oneLevel[i] = curr->val;
                }

                if(curr->left != nullptr) {
                    q.push(curr->left);
                }
                if (curr->right != nullptr) {
                    q.push(curr->right);
                }
            }
            reverse = !reverse;
            zigzag.push_back(oneLevel);
        }
        
        return zigzag;        
    }
};

## Time Complexity
$O(n)$
$n$ is the number of nodes in the n-ary tree referred by `root`.
## Space Complexity
$O(n)$
$n$ is the number of nodes in the n-ary tree referred by `root`.
# Miscellane
<!--
# Test Cases

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

[1]

[]

[1,2,3,4,null,null,5]

[1,null,2,3]

[4,2,7,1,3]

[18,2,22,null,null,null,63,null,84,null,null]

[18,6,22,5,7,11,63,1,2,9,8,10,24,12,84]

-->