【题目描述】
给你二叉树的根节点 root
,返回其节点值 自底向上的层序遍历 。 (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)
https://leetcode.cn/problems/binary-tree-level-order-traversal-ii/
【示例】
【代码】admin
在 102.二叉树的层序遍历 基础上把数据反转即可
// 2023-1-13
package com.company;
import java.util.*;
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 {
List<List<Integer>> res = new ArrayList<>();
public List<List<Integer>> levelOrderBottom(TreeNode root) {
check(root);
return res;
}
private void check(TreeNode root) {
Queue<TreeNode> queue = new LinkedList<>();
if (root == null) return;
queue.add(root);
while (!queue.isEmpty()){
List<Integer> list = new ArrayList<>();
int len = queue.size();
while (len >0){
TreeNode node = queue.poll();
list.add(node.val);
if (node.left != null) queue.offer(node.left);
if (node.right != null) queue.offer(node.right);
len--;
}
res.add(list);
}
Collections.reverse(res) ;
}
}
public class Test {
public static void main(String[] args) {
}
}
【代码】代码随想录
// 107. 二叉树的层序遍历 II
public class N0107 {
/**
* 解法:队列,迭代。
* 层序遍历,再翻转数组即可。
*/
public List<List<Integer>> solution1(TreeNode root) {
List<List<Integer>> list = new ArrayList<>();
Deque<TreeNode> que = new LinkedList<>();
if (root == null) {
return list;
}
que.offerLast(root);
while (!que.isEmpty()) {
List<Integer> levelList = new ArrayList<>();
int levelSize = que.size();
for (int i = 0; i < levelSize; i++) {
TreeNode peek = que.peekFirst();
levelList.add(que.pollFirst().val);
if (peek.left != null) {
que.offerLast(peek.left);
}
if (peek.right != null) {
que.offerLast(peek.right);
}
}
list.add(levelList);
}
List<List<Integer>> result = new ArrayList<>();
for (int i = list.size() - 1; i >= 0; i-- ) {
result.add(list.get(i));
}
return result;
}
}