103.binary-tree-zigzag-level-order-traversal.md

## 题目地址
https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/description/

## 题目描述
和leetcode 102 基本是一样的，思路是完全一样的。

```
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

For example:
Given binary tree [3,9,20,null,null,15,7],
    3
   / \
  9  20
    /  \
   15   7
return its zigzag level order traversal as:
[
  [3],
  [20,9],
  [15,7]
]
```

## 思路

这道题可以借助`队列`实现，首先把root入队，然后入队一个特殊元素Null(来表示每层的结束)。


然后就是while(queue.length), 每次处理一个节点，都将其子节点（在这里是left和right）放到队列中。

然后不断的出队， 如果出队的是null，则表式这一层已经结束了，我们就继续push一个null。


## 关键点解析

- 队列

- 队列中用Null(一个特殊元素)来划分每层

- 树的基本操作- 遍历 - 层次遍历（BFS）


## 代码

```js
/*
 * @lc app=leetcode id=103 lang=javascript
 *
 * [103] Binary Tree Zigzag Level Order Traversal
 *
 * https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/description/
 *
 * algorithms
 * Medium (40.57%)
 * Total Accepted:    201.2K
 * Total Submissions: 493.7K
 * Testcase Example:  '[3,9,20,null,null,15,7]'
 *
 * Given a binary tree, return the zigzag level order traversal of its nodes'
 * values. (ie, from left to right, then right to left for the next level and
 * alternate between).
 * 
 * 
 * For example:
 * Given binary tree [3,9,20,null,null,15,7],
 * 
 * ⁠   3
 * ⁠  / \
 * ⁠ 9  20
 * ⁠   /  \
 * ⁠  15   7
 * 
 * 
 * 
 * return its zigzag level order traversal as:
 * 
 * [
 * ⁠ [3],
 * ⁠ [20,9],
 * ⁠ [15,7]
 * ]
 * 
 * 
 */
/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var zigzagLevelOrder = function(root) {
  if (!root) return [];   
  const items = [];
  let isOdd = true;
  let levelNodes = [];
  
  const queue = [root, null];


  while(queue.length > 0) {
      const t = queue.shift();

      if (t) {
          levelNodes.push(t.val)
          if (t.left) {
            queue.push(t.left)
          }
          if (t.right) {
            queue.push(t.right)
          }
      } else {
        if (!isOdd) {
          levelNodes = levelNodes.reverse();
        }
        items.push(levelNodes)
        levelNodes = [];
        isOdd = !isOdd;
        if (queue.length > 0) {
            queue.push(null);
        }
      }
  }

  return items
    
};
```