SOJ 1310. Right-Heavy Tree
2013-08-26 00:21
246 查看
Description
A right-heavy tree is a binary tree where the value of a node is greater than or equal to the values of the nodes in its left subtree and less than the values of the nodes in its right subtree. A right-heavy tree could be empty.
Write a program that will create a right-heavy tree from a given input sequence and then traverse the tree and printing the value of the node each time a node is visited using inorder, preorder and postorder traversal.
The program should create the nodes in the tree dynamically. Thus, basically the tree can be of any size limited only by the amount of memory available in the computer.
Input
The input will contain several test cases, each of them as described below.
The first number in the input indicates the number of nodes in the tree. Then, the input is followed by the integers comprising the values of the nodes of the tree.
Output
For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line. The output will be the sequence of node and labeled by the traversal method used, as shown in the sample output below.
Sample Input
Copy sample input to clipboard
Sample Output
Hint
The number of vertexs is no more than 200000.
A right-heavy tree is a binary tree where the value of a node is greater than or equal to the values of the nodes in its left subtree and less than the values of the nodes in its right subtree. A right-heavy tree could be empty.
Write a program that will create a right-heavy tree from a given input sequence and then traverse the tree and printing the value of the node each time a node is visited using inorder, preorder and postorder traversal.
The program should create the nodes in the tree dynamically. Thus, basically the tree can be of any size limited only by the amount of memory available in the computer.
Input
The input will contain several test cases, each of them as described below.
The first number in the input indicates the number of nodes in the tree. Then, the input is followed by the integers comprising the values of the nodes of the tree.
Output
For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line. The output will be the sequence of node and labeled by the traversal method used, as shown in the sample output below.
Sample Input
Copy sample input to clipboard
8 8 2 4 7 5 3 1 6 9 5 5 6 3 2 9 3 3 2 8 4 2 1 4 3 2 5 1 0
Sample Output
Inorder: 1 2 3 4 5 6 7 8 Preorder: 8 2 1 4 3 7 5 6 Postorder: 1 3 6 5 7 4 2 8 Inorder: 2 2 3 3 3 5 5 6 9 Preorder: 5 5 3 2 2 3 3 6 9 Postorder: 2 3 3 2 3 5 9 6 5 Inorder: 1 1 2 2 3 4 4 5 Preorder: 4 2 1 1 2 4 3 5 Postorder: 1 2 1 3 4 2 5 4 Inorder: Preorder: Postorder:
Hint
The number of vertexs is no more than 200000.
#include<iostream> #include<memory.h> using namespace std; struct BT{ struct RHT { int data; RHT *left; RHT *right; RHT(int a,RHT *left=NULL, RHT *RHT=NULL):data(a),left(left),right(right) {} }; RHT *root; BT() { root=NULL; } void insert(const int & dat ) { return ins(root,dat); } void ins(RHT * & root,const int & dat) { if(root != NULL) { if(root->data>=dat) ins(root->left,dat); else ins(root->right,dat); } else root = new RHT(dat);//for pointer's ,you must new operator } void Inorder(RHT * &root) { if(root!=NULL) { Inorder(root->left); cout<<" "<<root->data; Inorder(root->right); } } void Preorder(RHT * &root) { if(root!=NULL) { cout<<" "<<root->data; Preorder(root->left); Preorder(root->right); } } void Postorder(RHT * &root) { if(root!=NULL) { Postorder(root->left); Postorder(root->right); cout<<" "<<root->data; } } }; int main() { int n; bool first=true; int temp=0; while(cin>>n) { if(!first) cout<<endl; first=false; BT bt; for (int i = 0; i < n; ++i) { cin>>temp; bt.insert(temp); } cout<<"Inorder:"; bt.Inorder(bt.root); cout<<endl; cout<<"Preorder:"; bt.Preorder(bt.root); cout<<endl; cout<<"Postorder:"; bt.Postorder(bt.root); cout<<endl; } return 0; },这一个版本用了0.79sec,之前写了一个版本的,可能是因为没有按地址传递,所以TLE了。构造树和前序中序后序遍历其实挺容易的。
相关文章推荐
- Sicily 1310. Right-Heavy Tree
- Sicily 1310. Right-Heavy Tree
- sicily 1310. Right-Heavy Tree
- 1310. Right-Heavy Tree
- soj 1814
- soj 3102 不明白
- soj 2932:道路
- SOJ 4139 APPLE 树状DP
- SOJ 4243: Saving Girl
- soj 1396. Q
- soj 1621. Circles in Circle
- soj - 1014 - Specialized Four-Dig
- soj 1939. 煎饼堆
- 1001. Black Magic (soj)
- Sicily 1142 排序 (SOJ 1142) 【搜索剪枝】
- SOJ 1034. Forest
- soj dsa extra 1002 sudoku
- soj 1698 Hungry Cow_三角函数
- 【异或】SOJ Find Differences
- soj题目分类