# 「力扣」第 107 题：二叉树的层次遍历 II（中等）
# 题目描述
Given the root of a binary tree, return the bottom-up level order traversal of its nodes' values. (i.e., from left to right, level by level from leaf to root).
Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: [[15,7],[9,20],[3]]
Example 2:
Input: root = [1]
Output: [[1]]
Example 3:
Input: root = []
Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 2000]. -1000 <= Node.val <= 1000
参考代码：
Java 代码：
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
public class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> res = new LinkedList<>();
if (root == null) {
return res;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
int size = queue.size();
List<Integer> curLevel = new ArrayList<>();
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
curLevel.add(node.val);
if (node.left != null) {
queue.add(node.left);
}
if (node.right != null) {
queue.add(node.right);
}
}
res.add(0, curLevel);
}
return res;
}
}
Python 代码：
from typing import List
from collections import deque
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution:
def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
res = []
if root is None:
return res
queue = deque()
queue.append(root)
while queue:
size = len(queue)
cur = []
for _ in range(size):
top = queue.popleft()
cur.append(top.val)
if top.left:
queue.append(top.left)
if top.right:
queue.append(top.right)
res.insert(0, cur)
return res
作者：liweiwei1419 链接：https://suanfa8.com/tree/solutions/0107-binary-tree-level-order-traversal-ii 来源：算法吧 著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。