PAT A1020 Tree Traversals [binary tree traversal]

Title Description

Link
after a given sequence and order, seeking layer sequence results

analysis

• Recursive, presented herein after the sequence and the first order, in order to seek
• Recursive: Consider the middle of the process
  • Preorder \ ([preL, preR] \ ) in sequence \ ([inL, inR] \ )
  • Root: \ (preL \)
  • In order root: \ (K, A [K] == A [preL] \) traversing obtained
  • The number of nodes in the left subtree sequence: \ (K-INL \) , the left subtree sequence \ ([inL, k-1 ] \)
  • Sequence preorder left subtree \ ([preL + 1, preL + k-inL] \)
  • Sequence right subtree of the number of nodes: \ (inR-K \) , the right subtree sequence \ ([k + 1, inR ] \)
  • Sequence preorder right subtree: \ ([K + preL-INL +. 1, PreR] \)
  • State function \ ((preL, preR, inL , inR) \)
• Terminator: sequence length of less than or equal to 0, i.e. \ (preR-preL <0 \ )
• Inside the concrete operation: each recursive, get the value of the root node, you can generate a root node, specifically how to hang it? The return value of the function can be written as a pointer and then back when you can hang up. The last return is the root of the entire tree

#include<bits/stdc++.h>
using namespace std;

struct node{
    int data;
    node *lchild;
    node *rchild;
};

const int maxn = 100;
int post[maxn], in[maxn];
node *create(int postL, int postR, int inL, int inR){
    if(postL > postR) return NULL;
    int k;
    for(k=inL;k<=inR;k++){
        if(in[k] == post[postR]) break;
    }
    int cnt = k-inL;
    node *root = new node;
    root->data = post[postR];
    root->lchild = create(postL, postL+cnt-1, inL, k-1);
    root->rchild = create(postL+cnt, postR-1, k+1, inR);
    return root;
}

vector<int> ans;
void layerOrder(node *root){
    queue<node*> q;
    q.push(root);
    while(!q.empty()){
        node *now = q.front();
        ans.push_back(now->data);
        q.pop();
        if(now->lchild) q.push(now->lchild);
        if(now->rchild) q.push(now->rchild);
    }
}

int main(){
    int n;
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>post[i];
    }
    for(int i=0;i<n;i++){
        cin>>in[i];
    }
    node *root = create(0, n-1, 0, n-1);
    layerOrder(root);
    for(int i=0;i<ans.size();i++){
        if(i==0) cout<<ans[i];
        else cout<<" "<<ans[i];
    }
    cout<<endl;

}