742. Closest Leaf in a Binary Tree查找最近的叶子节点
[抄题]:
Given a binary tree where every node has a unique value, and a target key k
, find the value of the nearest leaf node to target k
in the tree.
Here, nearest to a leaf means the least number of edges travelled on the binary tree to reach any leaf of the tree. Also, a node is called a leaf if it has no children.
In the following examples, the input tree is represented in flattened form row by row. The actual root
tree given will be a TreeNode object.
Example 1:
Input:
root = [1, 3, 2], k = 1
Diagram of binary tree:
1
/ \
3 2 Output: 2 (or 3) Explanation: Either 2 or 3 is the nearest leaf node to the target of 1.
Example 2:
Input:
root = [1], k = 1
Output: 1 Explanation: The nearest leaf node is the root node itself.
Example 3:
Input:
root = [1,2,3,4,null,null,null,5,null,6], k = 2
Diagram of binary tree:
1
/ \
2 3
/
4
/
5
/
6 Output: 3
Explanation: The leaf node with value 3 (and not the leaf node with value 6) is nearest to the node with value 2.
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
不知道还要找点,把路径存在hashmap中
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[一句话思路]:
bfs的过程中,cur节点的左、右、map中存储的路径都要放进q
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
- dfs函数的作用是返回有效的 值为k的节点,所以结果是左右节点的时候也需要返回
- 左右节点均为空的时候,再返回root.val的数值
[二刷]:
- dfs函数中,先把左节点放进去,再返回整个的left
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
用map存路径map.put(root.left, root);,然后用dfs找到k
[复杂度]:Time complexity: O(n) Space complexity: O(n)
[算法思想:迭代/递归/分治/贪心]:
[关键模板化代码]:
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
[是否头一次写此类driver funcion的代码] :
[潜台词] :
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int findClosestLeaf(TreeNode root, int k) {
//corner case
if (root == null) return -1; //initialiazation: map, q
//put first node into q, add left, right, route
HashMap<TreeNode, TreeNode> map = new HashMap<TreeNode, TreeNode>();
PriorityQueue<TreeNode> q = new PriorityQueue<TreeNode>();
TreeNode match = dfsTree(k, root, map); q.add(match);
while (!q.isEmpty()) {
TreeNode cur = q.poll();
if (cur.left == null && cur.right == null) return cur.val;
if (cur.left != null) q.add(cur.left);
if (cur.right != null) q.add(cur.right);
if (map.containsKey(cur)) {
q.add(map.get(cur));
map.remove(cur);
}
} return -1;
} public TreeNode dfsTree(int k, TreeNode root, Map<TreeNode, TreeNode> map) {
//corner case : null
if (root == null) return null; //return if left & right is null
if (root.val == k) return root; //put left into map and return left
if (root.left != null) {
map.put(root.left, root);
TreeNode left = dfsTree(k, root.left, map);
if (left != null) return left;
} //put left into map and return left
if (root.right != null) {
map.put(root.right, root);
TreeNode right = dfsTree(k, root.right, map);
if (right != null) return right;
} //return null
return null;
}
}
742. Closest Leaf in a Binary Tree查找最近的叶子节点的更多相关文章
- LeetCode 742. Closest Leaf in a Binary Tree
原题链接在这里:https://leetcode.com/problems/closest-leaf-in-a-binary-tree/ 题目: Given a binary tree where e ...
- [LeetCode] Closest Leaf in a Binary Tree 二叉树中最近的叶结点
Given a binary tree where every node has a unique value, and a target key k, find the value of the n ...
- Leetcode: Closest Leaf in a Binary Tree
Given a binary tree where every node has a unique value, and a target key k, find the value of the n ...
- jquery zTree 查找所有的叶子节点
jquery zTree 查找所有的叶子节点 // 保存所有叶子节点 10 为初始化大小,并非数组上限 var arrayObj = new Array([10]); /* treeNode: 根节点 ...
- [LeetCode] Find Leaves of Binary Tree 找二叉树的叶节点
Given a binary tree, find all leaves and then remove those leaves. Then repeat the previous steps un ...
- [LeetCode] 366. Find Leaves of Binary Tree 找二叉树的叶节点
Given a binary tree, find all leaves and then remove those leaves. Then repeat the previous steps un ...
- easyui tree 判断是否是叶子节点
<input class="add" id="add" style="display: none" type="submit ...
- [leetcode]236. Lowest Common Ancestor of a Binary Tree 二叉树最低公共父节点
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree. According ...
- [LeetCode] 298. Binary Tree Longest Consecutive Sequence 二叉树最长连续序列
Given a binary tree, find the length of the longest consecutive sequence path. The path refers to an ...
随机推荐
- 如何将项目连接数据库(连接mysql)
首先需要在项目中加入这一串代码: //加载驱动类 连接数据库有多种方式 比如:jdbc 桥接 Connection con=null; try { Class.forName("com.my ...
- mongodb分片balance
查看balance状态 mongos> sh.getBalancerState()true 通过balance锁查看balance活动 如果state是2,表示balance锁已经被获取 m ...
- 十九、springcloud(五)配置中心本地示例和refresh
1.创建spring-cloud-houge-config子项目,测试需要的项目入下 2.添加依赖 <dependency> <groupId>org.springframew ...
- InnoDB存储引擎文件
InnoDB存储引擎文件 MySQL数据库包括数据库本身的文件和存储引擎文件.数据库自身的文件由参数文件(my.cnf).错误日志文件.慢查询日志文件.查询日志文件.二进制日志文件.套接字文件.pid ...
- PhpAdmin支持登录远程数据库服务器
转载:http://www.cnblogs.com/andydao/p/4227312.html 该数据,百度搜不到,Google1分钟搞定 一.如何设置phpMyAdmin自动登录? 首先在根目录找 ...
- 关于音频总线IIS的学习---Verilog
关于音频总线IIS的学习---Verilog 主要思想: 在分析寄存器的值变化的时候,将时钟的边沿分两边来看,边沿之前,边沿之后,在always 块语句里面用来分析判断的寄存器的值,都应该用边沿变化之 ...
- 秒懂,Java 注解 (Annotation)你可以这样学 - CSDN博客
https://blog.csdn.net/briblue/article/details/73824058 文章开头先引入一处图片. 这处图片引自老罗的博客.为了避免不必要的麻烦,首先声明我个人比较 ...
- Response的Content-Type一览
文件扩展名 Content-Type(Mime-Type) 文件扩展名 Content-Type(Mime-Type) .* application/octet-stream .tif image/t ...
- python:带参数的装饰器,函数的有用信息
一.带参数的装饰器,函数的有用信息 def func1(): '''此函数的功能是完成的登陆的功能 return: 返回值是登陆成功与否(true,false) ''' print(333) func ...
- JVM学习总结(一):Java内存区域
一.JVM运行时数据区 1.程序计数器: (1)一块较小的线程私有的内存空间. (2)JVM的多线程是通过线程轮流切换并分配处理器执行时间的方式来实现的,在任何一个确定的时刻,一个处理器(或一个内核) ...