PAT 甲级 1020 Tree Traversals (25 分)（二叉树已知后序和中序建树求层序）

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

这个推了蛮久的。。。。。不太熟练

    //左子树 
    if(k-l2==0){
        tree[root].l = -1;//左子树为空
    }
    else{
        buildTree(2*root,l1,l1+k-l2-1,l2,k-1);    
    }
    //右子树 
    if(k+1>r2){
        tree[root].r = -1;//右子树为空
    }
    else{
        buildTree(2*root+1,l1+k-l2,r1-1,k+1,r2);
    }

AC代码：

#include<bits/stdc++.h>
using namespace std;
int post[35];
int in[35];
int n;
struct node{
    int v;
    int l;
    int r;
}tree[2005];
queue<int>q;
void buildTree(int root,int l1,int r1,int l2,int r2){//后序,中序 
    //cout<<root<<" "<<l1<<"-"<<r1<<" "<<l2<<"-"<<r2<<endl;
    //先找根节点
    tree[root].v = post[r1];
    if(l1==r1){    //只有一个节点    
        tree[root].l = -1;
        tree[root].r = -1;
        return;
    }else{
        tree[root].l = 2*root;
        tree[root].r = 2*root+1;
    }
    //找一下根在中序上的位置
    int k;
    for(int i=l2;i<=r2;i++){
        if(in[i]==post[r1]){
            k=i;//前面k-l2个数就是左子树 
            break;
        }
    } 
    //左子树 
    if(k-l2==0){
        tree[root].l = -1;//左子树为空
    }
    else{
        buildTree(2*root,l1,l1+k-l2-1,l2,k-1);    
    }
    //右子树 
    if(k+1>r2){
        tree[root].r = -1;//右子树为空
    }
    else{
        buildTree(2*root+1,l1+k-l2,r1-1,k+1,r2);
    }
}
int main(){
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>post[i];
    }
    for(int i=1;i<=n;i++){
        cin>>in[i];
    }
    buildTree(1,1,n,1,n);
    //bfs 层序 
    while(!q.empty()) q.pop();
    q.push(1);
    while(!q.empty()){
        node x=tree[q.front()];
        q.pop();
        cout<<x.v;
        if(x.l!=-1){
            q.push(x.l);
        }
        if(x.r!=-1){
            q.push(x.r);
        }
        if(!q.empty()){
            cout<<" ";
        }
    }
    return 0;
}