---
comments: true
difficulty: 中等
edit_url: https://github.com/doocs/leetcode/edit/main/solution/0100-0199/0107.Binary%20Tree%20Level%20Order%20Traversal%20II/README.md
tags:
- 树
- 广度优先搜索
- 二叉树
---
<!-- problem:start -->
# [107. 二叉树的层序遍历 II](https://leetcode.cn/problems/binary-tree-level-order-traversal-ii)
[English Version](/solution/0100-0199/0107.Binary%20Tree%20Level%20Order%20Traversal%20II/README_EN.md)
## 题目描述
<!-- description:start -->
<p>给你二叉树的根节点 <code>root</code> ，返回其节点值 <strong>自底向上的层序遍历</strong> 。 （即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0107.Binary%20Tree%20Level%20Order%20Traversal%20II/images/tree1.jpg" style="width: 277px; height: 302px;" />
<pre>
<strong>输入：</strong>root = [3,9,20,null,null,15,7]
<strong>输出：</strong>[[15,7],[9,20],[3]]
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>root = [1]
<strong>输出：</strong>[[1]]
</pre>
<p><strong>示例 3：</strong></p>
<pre>
<strong>输入：</strong>root = []
<strong>输出：</strong>[]
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li>树中节点数目在范围 <code>[0, 2000]</code> 内</li>
<li><code>-1000 <= Node.val <= 1000</code></li>
</ul>
<!-- description:end -->
## 解法
<!-- solution:start -->
### 方法一：BFS
我们可以使用 BFS 的方法来解决这道题。首先将根节点入队，然后不断地进行以下操作，直到队列为空：
- 遍历当前队列中的所有节点，将它们的值存储到一个临时数组 $t$ 中，然后将它们的孩子节点入队。
- 将临时数组 $t$ 存储到答案数组中。
最后将答案数组反转后返回即可。
时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点个数。
<!-- tabs:start -->
#### Python3
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def levelOrderBottom(self, root: Optional[TreeNode]) -> List[List[int]]:
ans = []
if root is None:
return ans
q = deque([root])
while q:
t = []
for _ in range(len(q)):
node = q.popleft()
t.append(node.val)
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
ans.append(t)
return ans[::-1]
```
#### Java
```java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
LinkedList<List<Integer>> ans = new LinkedList<>();
if (root == null) {
return ans;
}
Deque<TreeNode> q = new LinkedList<>();
q.offerLast(root);
while (!q.isEmpty()) {
List<Integer> t = new ArrayList<>();
for (int i = q.size(); i > 0; --i) {
TreeNode node = q.pollFirst();
t.add(node.val);
if (node.left != null) {
q.offerLast(node.left);
}
if (node.right != null) {
q.offerLast(node.right);
}
}
ans.addFirst(t);
}
return ans;
}
}
```
#### C++
```cpp
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> ans;
if (!root) {
return ans;
}
queue<TreeNode*> q{{root}};
while (!q.empty()) {
vector<int> t;
for (int n = q.size(); n; --n) {
auto node = q.front();
q.pop();
t.push_back(node->val);
if (node->left) {
q.push(node->left);
}
if (node->right) {
q.push(node->right);
}
}
ans.push_back(t);
}
reverse(ans.begin(), ans.end());
return ans;
}
};
```
#### Go
```go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func levelOrderBottom(root *TreeNode) (ans [][]int) {
if root == nil {
return
}
q := []*TreeNode{root}
for len(q) > 0 {
t := []int{}
for n := len(q); n > 0; n-- {
node := q[0]
q = q[1:]
t = append(t, node.Val)
if node.Left != nil {
q = append(q, node.Left)
}
if node.Right != nil {
q = append(q, node.Right)
}
}
ans = append(ans, t)
}
for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
ans[i], ans[j] = ans[j], ans[i]
}
return
}
```
#### TypeScript
```ts
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function levelOrderBottom(root: TreeNode | null): number[][] {
const ans: number[][] = [];
if (!root) {
return ans;
}
const q: TreeNode[] = [root];
while (q.length) {
const t: number[] = [];
const qq: TreeNode[] = [];
for (const { val, left, right } of q) {
t.push(val);
left && qq.push(left);
right && qq.push(right);
}
ans.push(t);
q.splice(0, q.length, ...qq);
}
return ans.reverse();
}
```
#### Rust
```rust
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::{cell::RefCell, collections::VecDeque, rc::Rc};
impl Solution {
pub fn level_order_bottom(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
let mut ans = Vec::new();
if let Some(root_node) = root {
let mut q = VecDeque::new();
q.push_back(root_node);
while !q.is_empty() {
let mut t = Vec::new();
for _ in 0..q.len() {
if let Some(node) = q.pop_front() {
let node_ref = node.borrow();
t.push(node_ref.val);
if let Some(ref left) = node_ref.left {
q.push_back(Rc::clone(left));
}
if let Some(ref right) = node_ref.right {
q.push_back(Rc::clone(right));
}
}
}
ans.push(t);
}
}
ans.reverse();
ans
}
}
```
#### JavaScript
```js
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[][]}
*/
var levelOrderBottom = function (root) {
const ans = [];
if (!root) {
return ans;
}
const q = [root];
while (q.length) {
const t = [];
const qq = [];
for (const { val, left, right } of q) {
t.push(val);
left && qq.push(left);
right && qq.push(right);
}
ans.push(t);
q.splice(0, q.length, ...qq);
}
return ans.reverse();
};
```
<!-- tabs:end -->
<!-- solution:end -->
<!-- problem:end -->