PAT 甲级 1020 Tree Traversals (25分)（后序中序链表建树，求层序）***重点复习

1020 Tree Traversals (25分)

 

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 (≤), 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

题意：

后序中序链表建树，求层序

AC代码：

#include<bits/stdc++.h>
using namespace std;
int n;
struct node{
    int data;
    node *left,*right;
};
int post[35];
int in[35];
vector<int>level;
queue<node*>q;
node *build(int pl,int pr,int il,int ir){
    if(pl>pr || il>ir) return NULL;
    int pos=-1;
    for(int i=il;i<=ir;i++){
        if(in[i]==post[pr]){
            pos=i;
            break;
        }
    }
    node *root=new node();
    root->data=post[pr];
    root->right=build(pr-(ir-pos),pr-1,pos+1,ir);//是ir-pos
    root->left=build(pl,pr-(ir-pos)-1,il,pos-1);
    return root;
}
int main(){
    cin>>n;
    for(int i=1;i<=n;i++) cin>>post[i];
    for(int i=1;i<=n;i++) cin>>in[i];
    node *root=build(1,n,1,n);
    q.push(root);
    while(!q.empty()){
        root=q.front();
        level.push_back(root->data);
        q.pop();
        if(root->left) q.push(root->left);
        if(root->right) q.push(root->right);
    }
    for(int i=0;i<level.size();i++){
        cout<<level.at(i);
        if(i!=level.size()-1) cout<<" ";
    }
    return 0;
}