「力扣」第 107 题:二叉树的层次遍历 II(中等)
传送门:107. 二叉树的层次遍历 II。
给定一个二叉树,返回其节点值自底向上的层次遍历。 (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)
例如:
给定二叉树[3,9,20,null,null,15,7]
,3 / \ 9 20 / \ 15 7
返回其自底向上的层次遍历为:
[ [15,7], [9,20], [3] ]
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