LeetCode - Easy - 107

]

[](()Analysis

方法一：BFS

方法二：DFS

[](()Submission

import java.util.ArrayList;

import java.util.Arrays;

import java.util.LinkedList;

import java.util.List;

import com.lun.util.BinaryTree.TreeNode;

public class BinaryTreeLevelOrderTraversalII {

//方法一：BFS

public List<List<Integer>> levelOrderBottom1(TreeNode root) {

List<List<Integer>> result = new LinkedList<>();

if (root == null)

return result;

TreeNode p = root;

LinkedList<TreeNode> queue = new LinkedList<>(Arrays.asList(p));

while (!queue.isEmpty()) {

int size = queue.size();

List<Integer> list = new ArrayList<>();

while (size-- > 0) {

p = queue.poll();

if (p.left != null) {

queue.offer(p.left);

}

if (p.right != null) {

queue.offer(p.right);

}

list.add(p.val);

}

result.add(0, list);

}

return result;

}

//方法二：DFS

public List<List<Integer>> levelOrderBottom2(TreeNode root) {

List<List<Integer>> wrapList = new LinkedList<List<Integer>>();

levelOrderBottom2(wrapList, root, 0);

return wrapList;

}

public void levelOrderBottom2(List<List<Integer>> list, TreeNode root, int level) {

if (root == null)

return;

if (level >= list.size()) {

list.add(0, new LinkedList<Integer>());

}

levelOrderBottom2(list, root.left, level + 1);

levelOrderBottom2(list, root.right, level + 1);

list.g 《一线大厂 Java 面试题解析+后端开发学习笔记+最新架构讲解视频+实战项目源码讲义》无偿开源 威信搜索公众号【编程进阶路】 et(list.size() - level - 1).add(root.val);

}

}

[](()Test

import static org.junit.Assert.*;

import static org.hamcrest.CoreMatchers.*;

import java.util.ArrayList;

import java.util.Arrays;

import java.util.List;