# LC 103. Binary Tree Zigzag Level Order Traversal ### [Problem link](https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/) ###### tags: `leedcode` `medium` `c++` `BFS` Given the <code>root</code> 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:** <img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/tree1.jpg" style="width: 277px; height: 302px;" /> ``` 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 <code>[0, 2000]</code>. - <code>-100 <= Node.val <= 100</code> ## Solution 1 - BFS #### C++ ```cpp= /** * 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) { vector<vector<int>> ans; bool isEven = false; queue<TreeNode *> q; if (root != nullptr) { q.push(root); } while (!q.empty()) { int sz = q.size(); vector<int> level; for (int i = 0; i < sz; i++) { TreeNode *node = q.front(); q.pop(); level.push_back(node->val); if (node->left) { q.push(node->left); } if (node->right) { q.push(node->right); } } if (isEven) { reverse(level.begin(), level.end()); } isEven = !isEven; ans.push_back(level); } return ans; } }; ``` >### Complexity >| | Time Complexity | Space Complexity | >| ----------- | --------------- | ---------------- | >| Solution 1 | O(n) | O(n) | ## Note x