102. Binary Tree Level Order Traversal

Recursion

/*
 * @lc app=leetcode id=102 lang=cpp
 *
 * [102] Binary Tree Level Order Traversal
 *
 * https://leetcode.com/problems/binary-tree-level-order-traversal/description/
 *
 * algorithms
 * Medium (68.36%)
 * Likes:    15464
 * Dislikes: 320
 * Total Accepted:    2.4M
 * Total Submissions: 3.5M
 * Testcase Example:  '[3,9,20,null,null,15,7]'
 *
 * Given the root of a binary tree, return the level order traversal of its
 * nodes' values. (i.e., from left to right, level by level).
 *
 *
 * Example 1:
 *
 *
 * Input: root = [3,9,20,null,null,15,7]
 * Output: [[3],[9,20],[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].
 * -1000 <= Node.val <= 1000
 *
 *
 */

// @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 {
private:
    vector<vector<int>> res;
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        if (!root) return res;
        dfs(root, 0);
        return res;
    }
    void dfs(TreeNode* root, int level) {
        if (res.size() == level) res.emplace_back(vector<int>{});
        res[level].emplace_back(root->val);
        if (root->left) {
            dfs(root->left, level + 1);
        }
        if (root->right) {
            dfs(root->right, level + 1);
        }
    }
};
// @lc code=end

• T: $O(N)$
• S: $O(N)$

Iteration

/**
 * 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>> levelOrder(TreeNode* root)
    {
        vector<vector<int>> res;
        queue<TreeNode*> q{{root}};
        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;
                nodesWithinLevel.push_back(node->val);
                q.push(node->left);
                q.push(node->right);
            }
            if (!nodesWithinLevel.empty()) res.push_back(nodesWithinLevel);
        }
        return res;
    }
};

• T: $O(N)$
• S: $O(N)$