<p>Given the <code>root</code> of a binary tree, return <em>the zigzag level order traversal of its nodes&#39; values</em>. (i.e., from left to right, then right to left for the next level and alternate between).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/tree1.jpg" style="width: 277px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [3,9,20,null,null,15,7]
<strong>Output:</strong> [[3],[20,9],[15,7]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [1]
<strong>Output:</strong> [[1]]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> root = []
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 2000]</code>.</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
</ul>

/**
 * 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)
        {
            return{};
        }
        queue<TreeNode*> q;
        vector<vector<int>> fin;
        q.push(root);
        int level=0;
        while(!q.empty())
        {
            int s=q.size();
            vector<int> ans;
            for(int i=0;i<s;i++)
            {
            TreeNode* node=q.front();
            q.pop();
            ans.push_back(node->val);
            if(node->left)
            {
                q.push(node->left);
            }
            if(node->right)
            {
                q.push(node->right);
            }
            }
            if(level%2==1)
            {
                reverse(ans.begin(),ans.end());
            }
            fin.push_back(ans);
            level++;
            }

        
        return fin;
        
    }
};
13300