# 103. Binary Tree Zigzag Level Order Traversal

## 1.問題

* 給予一個binary tree, 回傳一個zigzag level order traversal的node's序列 (例如, 從左到右, 從leaf到root的方向層層排列)

![](/files/-LH3-JbNo5kNGSJX4lsr)

## 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)與Binary Tree Level Order Traversal 的解法相同, 但是用一個flag來控制方向
* 向前放入元素

```
v.insert(v.begin(), value);
```

* 向後放入元素

```
 v.push_back(value);
```

## 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>> zigzagLevelOrder(TreeNode* root) {
        vector<vector<int>> res;
        levelOrder(res, root);
        return res;
    }
private:
    void levelOrder(vector<vector<int>>& res, TreeNode* root) {
        if (!root) {
            return;
        }
        
        queue<TreeNode*> q;
        q.push(root);
        int cnt = 1;
        while (!q.empty()) {
            int size = q.size();
            vector<int> record;
            for (int i = 0; i < size; i++) {
                TreeNode* front;
                front = q.front();
                q.pop();
                if (cnt % 2) {
                    record.push_back(front->val);
                } else {
                    record.insert(record.begin(), front->val);
                }
                
                if (front->left) {
                    q.push(front->left);
                }
                if (front->right) {
                    q.push(front->right);
                }   
            }
            cnt++;
            res.push_back(record);
        }
    }
 };
```

## 4.Performance

![](/files/-LH2y4kzOBY0JIj5VfdG)


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