> For the complete documentation index, see [llms.txt](https://imhuay.gitbook.io/studies/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/niu-ke-0014-zhong-deng-an-zhi-zi-xing-shun-xu-da-yin-er-cha-shu.md).

# 按之字形顺序打印二叉树

![last modify](https://img.shields.io/static/v1?label=last%20modify\&message=2022-10-14%2014%3A59%3A33\&color=yellowgreen\&style=flat-square) [![](https://img.shields.io/static/v1?label=\&message=%E4%B8%AD%E7%AD%89\&color=yellow\&style=flat-square)](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/pages/R5NyOzkn3qAZy7wCx1pS#中等) [![](https://img.shields.io/static/v1?label=\&message=%E7%89%9B%E5%AE%A2\&color=green\&style=flat-square)](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/pages/R5NyOzkn3qAZy7wCx1pS#牛客) [![](https://img.shields.io/static/v1?label=\&message=%E4%BA%8C%E5%8F%89%E6%A0%91/%E6%A0%91\&color=blue\&style=flat-square)](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/pages/R5NyOzkn3qAZy7wCx1pS#二叉树树) [![](https://img.shields.io/static/v1?label=\&message=%E6%A0%88/%E9%98%9F%E5%88%97\&color=blue\&style=flat-square)](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/pages/R5NyOzkn3qAZy7wCx1pS#栈队列)

**问题简述**

```
层序遍历二叉树，按之字形打印每层。
```

> [按之字形顺序打印二叉树\_牛客题霸\_牛客网](https://www.nowcoder.com/practice/91b69814117f4e8097390d107d2efbe0)

**思路**

* 队列 + 奇偶讨论，思路比较简单，因为需要把层分离，所以需要借助的辅助变量比较多，详见代码；

<details>

<summary><strong>Python</strong></summary>

```python
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param pRoot TreeNode类 
# @return int整型二维数组
#
class Solution:
    def Print(self , pRoot: TreeNode) -> List[List[int]]:
        # write code here
        if not pRoot: return []
        
        from collections import deque
        
        ret = []
        q = deque()
        q.append(pRoot)
        cnt = 1
        nxt = 0  # 下一层需要遍历的节点数
        lv = 0  # 已经遍历的层数
        tmp = []  # 当前层缓存的节点数
        while cnt:
            cnt -= 1
            node = q.popleft()
            tmp.append(node.val)
            
            if node.left:
                q.append(node.left)
                nxt += 1
            if node.right:
                q.append(node.right)
                nxt += 1
            
            if cnt == 0:
                if lv % 2:
                    ret.append(tmp[::-1])
                else:
                    ret.append(tmp)
                tmp = []
                lv += 1
                cnt = nxt
                nxt = 0
                
        return ret
```

</details>