# 102. Binary Tree Level Order Traversal

## 1.問題

* 給予一個binary tree, 回傳一個level order traversal的node's序列 (例如, 從左到右, 一層接一層)

![](https://901207480-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LGKoChvN9am4__HCIRK%2F-LH2YQTdpFVmQad9eBw5%2F-LH2YSHtHJI40yN6YftE%2F2018071002.jpg?alt=media\&token=68b0efa9-e09b-4ddf-a6f3-9f8c7f08d1ef)

## 2.想法

* 可用level-order traversal, 廣度優先搜尋(BFS)的概念:
  * 1.先將root node放入一開始的queue
  * 2.queue當前中的每一個元素若有子樹的, 則加入queue中 (等待下一次)

## 3.程式碼

```
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> v;
        if (!root) {
            return v;
        }
        queue<TreeNode*> q;
        vector<int> vTemp;
        vTemp.push_back(root->val);
        v.push_back(vTemp);
        q.push(root);
        
        while (!q.empty()){
            vector<int> vTemp;
            int size = q.size();
            for(int i = 0 ; i < size; i++){
                TreeNode* node = q.front();
                q.pop();
                if (node->left){
                    vTemp.push_back(node->left->val);
                    q.push(node->left);
                }
                if (node->right){
                    vTemp.push_back(node->right->val);
                    q.push(node->right);
                }
            }
            if (!vTemp.empty()){
                v.push_back(vTemp);
            }
        }
        return v;
    }
};
```

## 4.Performance

![](https://901207480-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LGKoChvN9am4__HCIRK%2F-LH2RJeZFO8zHbL75DHu%2F-LH2YA-boZ-aNH8ywx54%2F2018071001.jpg?alt=media\&token=378f0dda-88fe-4344-a037-21c8ac2f831f)