107. Binary Tree Level Order Traversal II
题目:
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
链接: http://leetcode.com/problems/binary-tree-level-order-traversal-ii/
题解:
这几道题都差不多,有关level order的。也可能有很好的方法,但是都没仔细想。基本思路和其他一样,就是BFS
Time Complexity - O(n), Space Complexity - O(1)。
/**
* Definition for a binary tree node.
* public 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 ArrayList<>();
if(root == null)
return res;
ArrayList<Integer> list = new ArrayList<>();
Queue<TreeNode> q = new LinkedList<>();
q.offer(root);
int curLevel = 1, nextLevel = 0; while(!q.isEmpty()) {
TreeNode node = q.poll();
curLevel--;
list.add(node.val);
if(node.left != null) {
q.offer(node.left);
nextLevel++;
}
if(node.right != null) {
q.offer(node.right);
nextLevel++;
}
if(curLevel == 0) {
curLevel = nextLevel;
nextLevel = 0;
res.add(0, new ArrayList<Integer>(list));
list.clear();
}
} return res;
}
}
二刷 :
方法跟一刷一样,就是跟普通level order traversal一样,但是保存结果的时候点到以下顺序。
Java:
Time Complexity - O(n), Space Complexity - O(1)。
/**
* Definition for a binary tree node.
* public 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 ArrayList<>();
if (root == null) {
return res;
}
Queue<TreeNode> q = new LinkedList<>();
q.offer(root);
int curLevel = 1, nextLevel = 0;
List<Integer> level = new ArrayList<>();
while (!q.isEmpty()) {
TreeNode node = q.poll();
curLevel--;
level.add(node.val);
if (node.left != null) {
q.offer(node.left);
nextLevel++;
}
if (node.right != null) {
q.offer(node.right);
nextLevel++;
}
if (curLevel == 0) {
curLevel = nextLevel;
nextLevel = 0;
res.add(0, new ArrayList<Integer>(level));
level.clear();
}
}
return res;
}
}
题外话: 2/14/2016
时间过得真的很快,不抓紧的话根本刷不完。前面一段时间效率极低,每天一两题,也不愿意花时间刷题,反而停留在舒适区,看看书,看看设计。后来返回来刷题发现根本不熟。
最后祝大家情人节快乐。
三刷:
再看了一下discuss,好像用dfs始终比bfs快.
Java:
/**
* Definition for a binary tree node.
* public 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 ArrayList<>();
if (root == null) {
return res;
}
List<Integer> level = new ArrayList<>();
Queue<TreeNode> q = new LinkedList<>();
q.offer(root);
int curLevel = 1, nextLevel = 0; while (!q.isEmpty()) {
TreeNode node = q.poll();
level.add(node.val);
curLevel--;
if (node.left != null) {
q.offer(node.left);
nextLevel++;
}
if (node.right != null) {
q.offer(node.right);
nextLevel++;
}
if (curLevel == 0) {
curLevel = nextLevel;
nextLevel = 0;
res.add(0, new ArrayList<Integer>(level));
level.clear();
}
}
return res;
}
}
Reference:
https://leetcode.com/discuss/5353/there-better-regular-level-order-traversal-reverse-result
https://leetcode.com/discuss/91100/share-java-2ms-recursive-solution
https://leetcode.com/discuss/22538/my-dfs-and-bfs-java-solution
https://leetcode.com/discuss/81189/java-1ms-beats-98%25-using-preorder
107. Binary Tree Level Order Traversal II的更多相关文章
- 102/107. Binary Tree Level Order Traversal/II
原文题目: 102. Binary Tree Level Order Traversal 107. Binary Tree Level Order Traversal II 读题: 102. 层序遍历 ...
- 【LeetCode】107. Binary Tree Level Order Traversal II (2 solutions)
Binary Tree Level Order Traversal II Given a binary tree, return the bottom-up level order traversal ...
- 【一天一道LeetCode】#107. Binary Tree Level Order Traversal II
一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 来源: htt ...
- (二叉树 BFS) leetcode 107. Binary Tree Level Order Traversal II
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left ...
- Java for LeetCode 107 Binary Tree Level Order Traversal II
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left ...
- LeetCode 107 Binary Tree Level Order Traversal II(二叉树的层级顺序遍历2)(*)
翻译 给定一个二叉树,返回从下往上遍历经过的每一个节点的值. 从左往右,从叶子到节点. 比如: 给定的二叉树是 {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 返回它从下 ...
- [LeetCode] 107. Binary Tree Level Order Traversal II 二叉树层序遍历 II
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left ...
- 剑指offer从上往下打印二叉树 、leetcode102. Binary Tree Level Order Traversal(即剑指把二叉树打印成多行、层序打印)、107. Binary Tree Level Order Traversal II 、103. Binary Tree Zigzag Level Order Traversal(剑指之字型打印)
从上往下打印二叉树这个是不分行的,用一个队列就可以实现 class Solution { public: vector<int> PrintFromTopToBottom(TreeNode ...
- [LeetCode]题解(python):107 Binary Tree Level Order Traversal II
题目来源 https://leetcode.com/problems/binary-tree-level-order-traversal-ii/ Given a binary tree, return ...
- leetcode 107 Binary Tree Level Order Traversal II ----- java
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left ...
随机推荐
- 笔试面试题-小米Git
题目描述: git是一种分布式代码管理工具,git通过树的形式记录文件的更改历史,比如: base'<--base<--A<--A' ^ | --- B<--B' 小米工程师常 ...
- 查看cpu、内存和硬盘
查看cpu cat /proc/cpuinfo 查看内存 top free -m 按兆为单位输出内存的已用,未用,总共等结果 cat /proc/meminfo |grep MemTotal 查看硬盘 ...
- Poj 3117 World Cup
1.Link: http://poj.org/problem?id=3117 2.Content: World Cup Time Limit: 1000MS Memory Limit: 65536 ...
- 《APUE》第五章笔记
第五章具体介绍了标准I/O库的各种细节,要是一一列出来,有费精力且可能列不全,故只讲平常多用到的.标准输入输出是由一大批函数组成的. 要记住,标准输入输出是有缓冲的,就是当缓冲区的数据满了的时候,才会 ...
- js数字格式化-四舍五入精简版
搜索网上的,数字格式化过余复杂,自己想了个简单方法,欢迎吐槽. 简化说明: '123333' => 12.3万 parseInt('123333') 字符串转整型 parseInt('12333 ...
- fill与memset的区别
fill 的头文件是<iostream> 命名空间是std: 在memset(a,0(-1),sizeof(a))全部初值定为0或-1时两者是没有多大区别; 但是在初值为其他值得时候就不同 ...
- 浏览器兼容CSS代码:按钮文字垂直居中(input button text vertical align)
经过测试的浏览器:IE6, IE7, IE8, IE9, Firefox, Chrome, Safiri, Maxthon 按钮的HTML代码: <input id="btn_comm ...
- 七、mysql索引选择
.myisam,bdb,innodb,memory 单表至少支持16个索引 .create index id_index on emp (id) 为emp表创建一个名为id_index的id字段的索引 ...
- mysql优化之sakila测试数据库
下载地址,选择相应的版本来进行安装测试 http://dev.mysql.com/doc/index-other.html 相关说明 http://dev.mysql.com/doc/sakila/e ...
- java运算符的优先级和详解
优先级 符号 名称 结合性(与操作数) 目数 说明 1 . 点 从左到右 双目 ( ) 圆括号 从左到右 [ ] 方括号 从左到右 2 + 正号 从右到左 单目 - 负号 从右到左 单目 ++ ...