1020 Tree Traversals (25point(s))

1020 Tree Traversals (25point(s))

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:
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7
Sample Output:
4 1 6 3 5 7 2

#include<bits/stdc++.h>
using namespace std;
const int maxn=35;
struct node{
    int data;
    node *lchild,*rchild;
};
int post[maxn],in[maxn];
node* create(int postL,int postR,int inL,int inR){
    if(postL>postR) return NULL;
    node* root=new node;
    root->data=post[postR];
    int k;
    for(k=inL;k<=inR;++k)
        if(in[k]==post[postR]) break;
    int leftNum=k-inL;
    root->lchild=create(postL,postL+leftNum-1,inL,k-1);
    root->rchild=create(postL+leftNum,postR-1,k+1,inR);
    return root;
}
void bfs(node* root){
    queue<node*> q;//存放地址
    q.push(root);
    int ans=0;
    while(!q.empty()){
        node* now=q.front();
        q.pop();
        if(ans!=0) printf(" ");
        ans++;
        printf("%d",now->data);
        if(now->lchild) q.push(now->lchild);
        if(now->rchild) q.push(now->rchild);
    }
    printf("\n");
}
int main(){
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;++i) scanf("%d",&post[i]);
    for(int i=0;i<n;++i) scanf("%d",&in[i]);
    node* root=create(0,n-1,0,n-1);
    bfs(root);
}