1020. Tree traversals (25)-pat question 1385: rebuilding a binary tree

1020. Tree traversals (25) Time Limit 400 ms memory limit 32000 kb code length limit 16000 B discriminant program standard author Chen, Yue

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

Input specification:

Each input file contains one test case. for each case, the first line gives a positive integer N (<= 30), the total number of nodes in the binary tree. the second line gives the postorder sequence and the third line gives the inorder sequence. all the numbers
In a line are separated by a space.

Output specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. all the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample input:

72 3 1 5 7 6 41 2 3 4 5 6 7

Sample output:

4 1 6 3 5 7 2

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Source: http://pat.zju.edu.cn/contests/pat-a-practise/1020

1. One method is to construct a tree using the back-order and Middle-order before BFs.

2. Another method is to calculate the left and right son nodes of each node without constructing a number.

A. left son: calculates the number of nodes with the right sub-number of the current node in the middle order, and then calculates the number of nodes with the right sub-tree in the current node position minus the number of nodes with the right sub-tree, position (in fact, the central order is to determine the size of the right subtree of the current node ).
rsize=right_size(root);left=hash_post[root]-rsize-1;
B. The right son traverses the previous node of the current node in the descending order. Right = hash_post [root]-1;
#include<iostream>#include<queue>#include<string.h>using namespace std;int *post,*visited;const int N=31;int hash_post[N],hash_in[N];int n;int right_size(int root){int i=0,size=0;for(i=hash_in[root]+1;i<=n-1;i++){if(visited[i]==false)size++;elsebreak;}return size;}int main(){    cin>>n;    post=new int[n];visited=new int [n];    int i,tmp;    for(i=0;i<n;i++){        cin>>post[i];hash_post[post[i]]=i;}    for(i=0;i<n;i++){cin>>tmp;hash_in[tmp]=i;}memset(visited,0,n*sizeof(int));queue<int> q;q.push(post[n-1]);int right=n-1;cout<<post[n-1];visited[hash_in[post[n-1]]]=true;while(!q.empty()){int root=q.front();int rsize=right_size(root);int left=hash_post[root]-rsize-1;//handle left sub tree rootif(left>=0&&visited[hash_in[post[left]]]==false){int left_root=post[left];cout<<" "<<left_root;visited[hash_in[left_root]]=true;q.push(left_root);}int right=hash_post[root]-1;//handle right sub tree rootif(right>=0&&visited[hash_in[post[right]]]==false){int right_root=post[right];cout<<" "<<right_root;visited[hash_in[right_root]]=true;q.push(right_root);}q.pop();}    return 0;}
9 degrees:
Topic 1385: rebuilding a binary tree

  
  

   
   
 
    Description: 
   
   
   
 
    
Enter the result of the forward and middle traversal of a binary tree. Create a New Binary Tree. Assume that the input results do not contain repeated numbers. For example, input pre-order traversal sequences {1, 2, 4, 7, 3, 5, 6, 8} and Middle-order traversal sequences {4, 7, 2, 1, 5, 3, 8, 6 }, then, the binary tree is rebuilt and its post-sequential traversal sequence is output.
 
    
 
    
 
   
  
  
  
  

   
   
 
    Input: 
   
   
   
 
    
The input may contain multiple test examples. For each test case,
 
    
The first input behavior is an integer N (1 <=n <= 1000): represents the number of nodes in a binary tree.
 
    
The second line of the input contains N integers (the range of each element a is (1 <= A <= 1000), which indicates the pre-order traversal sequence of the binary tree.
 
    
The third line of the input contains N integers (where the range of each element a is (1 <= A <= 1000): indicates the ordinal traversal sequence of the binary tree.
 
   
  
  
  
  

   
   
 
    Output: 
   
   
   
 
    
Output a line for each test case:
 
    
If the Pre-and mid-order traversal sequences given in the question can constitute a binary tree, N integers are output, representing the post-order traversal sequence of the binary tree. Each element is followed by spaces.
 
    
If the pre-order and mid-order traversal sequences given in the question cannot constitute a binary tree, "no" is output ".
 
   
  
  
  
  

   
   
 
    Sample input: 
   
   
   
 
    
81 2 4 7 3 5 6 84 7 2 1 5 3 8 681 2 4 7 3 5 6 84 1 2 7 5 3 8 6
 
   
  
  
  
  

   
   
 
    Sample output: 
   
   
   
 
    
7 4 2 5 8 6 3 1 No
 
   
  
  
Recommendation index :※※
Source: http://ac.jobdu.com/problem.php? PID = 1, 1385
This is the basic structure of the post-order based on the pre-order and Middle-order.
#include<iostream>using namespace std;typedef struct node{    int val;    node *left;    node *right;}node;bool flag;node * rebuild(int *pre,int *in ,int length){    node *root=new node();    root->val=pre[0];root->left=NULL;root->right=NULL;    int index;    for(index=0;index<length;index++)        if(in[index]==root->val)            break;if(index>=length){flag=true;return NULL;}if(index>0)root->left=rebuild(pre+1,in,index);if(length-index-1>0)root->right=rebuild(pre+index+1,in+index+1,length-index-1);return root;}void print_post(node* root){    if(root!=NULL){        print_post(root->left);        print_post(root->right);cout<<root->val<<" ";}}int main(){    int n,i;    while( cin>>n){flag=false;int *pre=new int[n];int *in=new int[n];for(i=0;i<n;i++)cin>>pre[i];for(i=0;i<n;i++)cin>>in[i];node *root;root=rebuild(pre,in,n);if(flag==true)cout<<"No"<<endl;else{print_post(root);cout<<endl;}    }    return 0;}