l2-011. Play Turn two fork tree

l2-011. Play two fork TreeTime limit MS
Memory Limit 65536 KB
Code length limit 8000 B
Chen, standard author of the procedure for the award of questions

Given a binary tree in the middle sequence traversal and pre-sequence traversal, please first make a mirror inversion of the tree, and then output the reverse sequence of sequential traversal. The so-called mirror reversal, refers to all non-leaf junction of the children to swap. This assumes that the key values are all non-equal positive integers.

Input Format:

Enter the first line to give a positive integer n (<=30), which is the number of nodes in the two-fork tree. The second line gives the sequential traversal sequence. The third line gives the sequence of its pre-sequence traversal. The numbers are separated by a space.

output Format:

Outputs the sequence of sequential traversal of the tree after it is inverted in a row. The numbers are separated by 1 spaces, with no extra spaces at the end of the line. Input Sample:

7
1 2 3 4 5 6 7
4 1 3 2 6 5 7

Sample output:

4 6 1 7 5 3 2

Submit Code

Analysis: After the completion of the tree directly in the BFS at the time around the exchange position.

#include <bits/stdc++.h> using namespace std;
int n;
const int maxn=35;
int BACK[MAXN];
int IN[MAXN];
    struct node {int data;
    node* Lchild;
node* Rchild;
};
    node* Solve (int inl,int inr,int backl,int backr) {if (Inl>inr | | backl>backr) return NULL;
    node* root=new node;
    int Gen=back[backl];
    root->data=gen;
    int record;
            for (int i=inl;i<=inr;i++) {if (In[i]==gen) {record=i;
        Break
    }} int numleft=record-inl;
    Root->lchild=solve (Inl,record-1,backl+1,backl+numleft);
    Root->rchild=solve (Record+1,inr,backl+numleft+1,backr);
return root;
} vector<int> ans;
    void BFs (node* root) {queue<node*> q;
    Q.push (root);
        while (!q.empty ()) {node* now=q.front ();
        Q.pop ();
        Ans.push_back (Now->data);
        if (now->rchild!=null) Q.push (now->rchild);

    if (now->lchild!=null) Q.push (now->lchild); }} void PRINT () {for (int i=0;i<ans.size (); i++) {cout<<ans[i];
    if (I!=ans.size ()-1) printf ("");
    }} int main () {cin>>n;
    for (int i=0;i<n;i++) {scanf ("%d", &in[i]);
    } for (int i=0;i<n;i++) {scanf ("%d", &back[i]);
    } BFS (Solve (0,n-1,0,n-1));


    Print ();
return 0;
 }