Populating Next Right Pointers in Each Node [LeetCode]
Given a binary tree
struct TreeLinkNode {
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *next;
}
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL
.
Initially, all next pointers are set to NULL
.
Note:
- You may only use constant extra space.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
For example,
Given the following perfect binary tree,
1
/ \
2 3
/ \ / \
4 5 6 7
After calling your function, the tree should look like:
1 -> NULL
/ \
2 -> 3 -> NULL
/ \ / \
4->5->6->7 -> NULL
Summary: The simplest way is BFS, but we need non-constant extra space. So, I traverses the tree Pre-order, and uses one node pointer for every level.
void traverse(TreeLinkNode *root, int depth, vector<TreeLinkNode *> ¤t) {
if(root->left != NULL && root->right != NULL){
if(current.size() < depth + )
current.push_back(root->right);
else{
current[depth]->next = root->left;
current[depth] = root->right;
}
root->left->next = root->right;
//traverse the left subtree
traverse(root->left, depth + , current);
//traverse the right subtree
traverse(root->right, depth + , current);
}
} void connect(TreeLinkNode *root) {
if(root == NULL)
return; vector<TreeLinkNode *> current;
traverse(root, , current);
}
Populating Next Right Pointers in Each Node [LeetCode]的更多相关文章
- Populating Next Right Pointers in Each Node leetcode java
题目: Given a binary tree struct TreeLinkNode { TreeLinkNode *left; TreeLinkNode *right; TreeLinkNode ...
- Leetcode 笔记 116 - Populating Next Right Pointers in Each Node
题目链接:Populating Next Right Pointers in Each Node | LeetCode OJ Given a binary tree struct TreeLinkNo ...
- Leetcode 笔记 117 - Populating Next Right Pointers in Each Node II
题目链接:Populating Next Right Pointers in Each Node II | LeetCode OJ Follow up for problem "Popula ...
- [LeetCode] Populating Next Right Pointers in Each Node II 每个节点的右向指针之二
Follow up for problem "Populating Next Right Pointers in Each Node". What if the given tre ...
- [LeetCode] Populating Next Right Pointers in Each Node 每个节点的右向指针
Given a binary tree struct TreeLinkNode { TreeLinkNode *left; TreeLinkNode *right; TreeLinkNode *nex ...
- LeetCode:Populating Next Right Pointers in Each Node I II
LeetCode:Populating Next Right Pointers in Each Node Given a binary tree struct TreeLinkNode { TreeL ...
- 【LeetCode OJ】Populating Next Right Pointers in Each Node II
Problem Link: http://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/ OK... ...
- leetcode@ [116/117] Populating Next Right Pointers in Each Node I & II (Tree, BFS)
https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/ Follow up for problem ...
- 【LeetCode】 Populating Next Right Pointers in Each Node 全然二叉树
题目:Populating Next Right Pointers in Each Node <span style="font-size:18px;">/* * Le ...
随机推荐
- Java_类文件及加载机制
类文件及类加载机制 标签(空格分隔): Java 本篇博客的重点是分析JVM是如何进行类的加载的,但同时我们会捎带着说一下Class类文件结构,以便对类加载机制有更深的理解. 类文件结构 平台无关性 ...
- Node.js 事件循环(Event Loop)介绍
Node.js 事件循环(Event Loop)介绍 JavaScript是一种单线程运行但又绝不会阻塞的语言,其实现非阻塞的关键是“事件循环”和“回调机制”.Node.js在JavaScript的基 ...
- emacs学习
(set-default-font "Consolas-14") // 设置字体及其大小 M-数字 命令 C-数字 命令
- 使用escape编码地址栏中的中文字符
在通过地址栏传递参数的时候,有时候会遇到中文参数,在获取这种中文参数值得时候, 往往会出现乱码, 解决办法如下: 在传递参数的使用 escape 函数进行编码,获取的时候再进行解码即可. 例如: va ...
- Xstream 学习地址
http://forestqqqq.iteye.com/category/301129
- MVC3远程验证
public class StudentModel { [Display(Name="学生编号")] public int StuId { set; get; } [Require ...
- poj3855Blast the Enemy!(多边形重心)
链接 #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> ...
- mysql delimiter
默认情况下,mysql遇到分号; 就认为是一个命令的终止符, 就会执行命令.而有些时候,我们不希望这样,比如存储过程中包含多个语句,这些语句以分号分割,我们希望这些语句作为一个命令,一起执行,怎么解决 ...
- gcc, numpy, rabbitmq等安装升级总结
1. 公司在下面目录安装了gcc-4.8.2,以支持c++11,可以通过在bashrc中添加来实现: PATH=/opt/compiler/gcc-4.8.2/bin:$PATH 2. 公司环境切换到 ...
- Android Fragment分页显示的实现
分页显示有两种方式 一种是使用ViewPager 另一种是使用FragmentTransaction 上代码 1 FragmentTransaction实现方式 public class MainAc ...