1115 Counting Nodes in a BST
题意:给出一棵二叉搜索树的插入序列,要求该树最后两层的结点个数。
思路:在树结点中增加一个数据域layer,表示该结点所在的层次。另外,设置数组level[]和变量maxLevel,level[i]表示第i层的结点个数,maxLevel表示树的最大层次,在层序遍历时更新即可。
代码:
#include <cstdio>
#include <queue>
using namespace std;
]={};
;
struct Node{
int val;
int layer;
Node *lchild,*rchild;
Node(),lchild(NULL),rchild(NULL){}
};
void insert(Node* &root,int val)
{
if(root==NULL) {
root=new Node(val);
return;
}
if(val<=root->val) insert(root->lchild,val);//根据题意,这里是<=,要仔细!
else insert(root->rchild,val);
}
void levelOrderTraversal(Node* root)
{
queue<Node*> q;
root->layer=;
q.push(root);
while(!q.empty()){
Node* pNode=q.front();
q.pop();
level[pNode->layer]++;
if(pNode->layer > maxLevel) maxLevel=pNode->layer;
if(pNode->lchild){
pNode->lchild->layer=pNode->layer+;
q.push(pNode->lchild);
}
if(pNode->rchild){
pNode->rchild->layer=pNode->layer+;
q.push(pNode->rchild);
}
}
}
int main()
{
int n,v;
scanf("%d",&n);
Node* root=NULL;
while(n--){
scanf("%d",&v);
insert(root,v);
}
levelOrderTraversal(root);
printf(],level[maxLevel-]+level[maxLevel]);
;
}
1115 Counting Nodes in a BST的更多相关文章
- PAT甲1115 Counting Nodes in a BST【dfs】
1115 Counting Nodes in a BST (30 分) A Binary Search Tree (BST) is recursively defined as a binary tr ...
- 1115 Counting Nodes in a BST (30 分)
1115 Counting Nodes in a BST (30 分) A Binary Search Tree (BST) is recursively defined as a binary tr ...
- [二叉查找树] 1115. Counting Nodes in a BST (30)
1115. Counting Nodes in a BST (30) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Y ...
- PAT 1115 Counting Nodes in a BST[构建BST]
1115 Counting Nodes in a BST(30 分) A Binary Search Tree (BST) is recursively defined as a binary tre ...
- PAT 甲级 1115 Counting Nodes in a BST
https://pintia.cn/problem-sets/994805342720868352/problems/994805355987451904 A Binary Search Tree ( ...
- 1115. Counting Nodes in a BST (30)
A Binary Search Tree (BST) is recursively defined as a binary tree which has the following propertie ...
- PAT 1115 Counting Nodes in a BST
A Binary Search Tree (BST) is recursively defined as a binary tree which has the following propertie ...
- PAT Advanced 1115 Counting Nodes in a BST (30) [⼆叉树的遍历,BFS,DFS]
题目 A Binary Search Tree (BST) is recursively defined as a binary tree which has the following proper ...
- PAT甲题题解-1115. Counting Nodes in a BST (30)-(构建二分搜索树+dfs)
题意:给出一个序列,构建二叉搜索树(BST),输出二叉搜索树最后两层的节点个数n1和n2,以及他们的和sum: n1 + n2 = sum 递归建树,然后再dfs求出最大层数,接着再dfs计算出最后两 ...
- PAT A 1115. Counting Nodes in a BST (30)【二叉排序树】
题目:二叉排序树,统计最后两层节点个数 思路:数组格式存储,insert建树,dfs遍历 #include<cstdio> #include<iostream> #includ ...
随机推荐
- BusyIndicator using MVVM 忙碌状态指示器的的实现
ViewModel 视图模型 public abstract class ViewModelBase : INotifyPropertyChanged { private bool isbusy; p ...
- CodeForces - 767C
花了6个小时,终于成功ac...... 两边dfs,第一遍求子树和,第二遍判断有没有2*t[s]/3和t[s]/3,因为要求的节点可能是在同一条线上,同时要有2*t[s]/3和t[s]/3的情况,且2 ...
- 解决loadrunner在脚本回放时长时间等待及在vugen中create controller scenario时报错的方法!超管用!!
解决loadrunner在脚本回放时长时间等待及在vugen中create controller scenario时报错的方法 经过咨询,有两种方法.经过实践,下面的方法1有效,方法2无效(我下载安装 ...
- 双系统在Linux下查看win的硬盘(Ubuntu 16.04 挂载Windows的 硬盘)
一般情况下,Linux的桌面系统能够直接查看到计算机各个硬盘的文件情况 但是,当我们想通过命令行查看Windows下的硬盘的时候,会发现在 /media/ (一般Windows下的盘会挂载到这里)文件 ...
- Git和SVN之间的区别
如果你在读这篇文章,说明你跟大多数开发者一样对GIT感兴趣,如果你还没有机会来试一试GIT,我想现在你就要了解它了. GIT不仅仅是个版本控制系统,它也是个内容管理系统(CMS),工作管理系统等.如果 ...
- I.MX6 U-Boot ping网络
/********************************************************************* * I.MX6 U-Boot ping网络 * 说明: * ...
- (十)java条件结构
条件结构 if(条件表达式) {}: if(条件表达式){} else {}; if(条件表达式){} else if(条件表达式) {} else if(条件表达式){} ...... else{} ...
- CodeForces - 622F:The Sum of the k-th Powers (拉格朗日插值法求自然数幂和)
There are well-known formulas: , , . Also mathematicians found similar formulas for higher degrees. ...
- POJ1160 Post Office (四边形不等式优化DP)
There is a straight highway with villages alongside the highway. The highway is represented as an in ...
- echarts.js:1136 Uncaught Error: Initialize failed: invalid dom.
一:错误描述:echarts.js:1136 Uncaught Error: Initialize failed: invalid dom. 二:错误原因:echarts在用json数据请求时未调用 ...