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# 102. Binary Tree Level Order Traversal

## 題目 :

Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).

### Example :

Note

Example 1:

Input: root = [3,9,20,null,null,15,7] Output: [[3],[9,20],[15,7]]

Example 2:

Input: root = [1] Output: [[1]]

Example 3:

Input: root = [] Output: []

#### 解題思路 :

• 這題其實就是用到queue的資料結構，並帶入廣度優先搜尋(BFS)的概念也就是廣義的level order traversal
• Graph常用到的bfs就是從tree的level order traversal推廣過去的
• 在binary tree這邊觀念很簡單就是一level為單位去遍歷
• 在搭配queue資料結構，當node從queue左邊出來時就把他的left childNode & right childNode從右邊放進去
• 並掌握每一round queue的長度，決定好每一round要拿出node的數量

### Queue解法

• time complexity: O(n) , space complexity: O(n)

Runtime: 0 ms, faster than 100.00% of Go online submissions for Binary Tree Level Order Traversal.

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 `````` ``````/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */ func levelOrder(root *TreeNode) [][]int { var queue []*TreeNode var res [][]int if root == nil { return res } queue = append(queue, root) for len(queue) != 0 { length := len(queue) tempArr := make([]int,0) for i := 0; i < length; i++ { node := queue[0] queue = queue[1:] if node.Left != nil { queue = append(queue, node.Left) } if node.Right != nil { queue = append(queue, node.Right) } tempArr = append(tempArr, node.Val) } res = append(res, tempArr) } return res } ``````