# Leetcode [No. 102] Binary Tree Level Order Traversal (MEDIUM)

## 題目

https://leetcode.com/problems/binary-tree-level-order-traversal/description/

## 思路
+ 這個解法其實就是用簡單的BFS來完成level order traversal.
+ 在每次pop之前都先去計算目前的queue有多少個element
+ 如此一來我們就可以知道在這次新增了多少個element

/**
 * 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>> ans;
        if(root == NULL) return ans;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty())
        {
            int levelSize = q.size(); // the size is queue size
            vector<int> vec;
            for(int i = 0 ;i < levelSize; i++)
            {
                TreeNode* current = q.front();
                vec.push_back(current->val);
                q.pop();
                if(current->left != NULL) q.push(current->left);
                if(current->right != NULL) q.push(current->right);
            }
            ans.push_back(vec);
        }
        return ans;
    }
};

### 解法分析
+ time complexity: O(n)

### 執行結果

---

## 二訪@2025/04/27

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        if (!root) return {};
        queue<TreeNode*> q;
        q.push(root);
        vector<vector<int>> res;
        while(!q.empty()){
            int size = q.size();
            vector<int> v(size);
            for (int i = 0 ; i < size; i++) {
                TreeNode* cur = q.front();
                q.pop();
                if (cur->left) q.push(cur->left);
                if (cur->right) q.push(cur->right);
                v[i] = cur->val;
            }
            res.emplace_back(v);
        }
        return res;
    }
};