103. Binary Tree Zigzag Level Order Traversal

/*
 * @lc app=leetcode id=103 lang=cpp
 *
 * [103] Binary Tree Zigzag Level Order Traversal
 *
 * https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/description/
 *
 * algorithms
 * Medium (60.32%)
 * Likes:    11317
 * Dislikes: 324
 * Total Accepted:    1.4M
 * Total Submissions: 2.3M
 * Testcase Example:  '[3,9,20,null,null,15,7]'
 *
 * 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
 *
 *
 */

// @lc code=start
/**
 * 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>> res;
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        if(!root) return res;
        levelOrder(root, 0);
        return res;
    }
    void levelOrder(TreeNode* root, int level) {
        if(!root) return;

        if(res.size() == level)
            res.push_back({});

        if(level % 2 == 0) {
            res[level].push_back(root->val);
        } else {
            res[level].insert(res[level].begin(), root->val);
        }
        levelOrder(root->left, 1 + level);
        levelOrder(root->right, 1 + level);
    }
};
// @lc code=end

• T: $O(N)$
• S: $O(N)$

class Solution {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root)
    {
        vector<vector<int>> res;
        queue<TreeNode*> q{{root}};
        bool leftToRight = true;
        while (!q.empty()) {
            vector<int> nodesWithinLevel;
            int qSize = q.size();
            for (int i = 0; i < qSize; i++) {
                TreeNode* node = q.front(); q.pop();
                if (!node) continue;
                if (leftToRight) {
                    nodesWithinLevel.push_back(node->val);
                } else {
                    nodesWithinLevel.insert(nodesWithinLevel.begin(), node->val);
                }
                q.push(node->left);
                q.push(node->right);
            }
            if (!nodesWithinLevel.empty()) {
                res.push_back(nodesWithinLevel);
                leftToRight = !leftToRight;
            }
        }
        return res;
    }
};

• T: $O(N)$
• S: $O(N)$