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
分析：根据后序遍历和中序遍历 输出层序遍历
我的做法是利用后序遍历和中序遍历建树 再层序遍历输出
利用两个变量来记录我们需要建子树的范围 另一个变量记录每层的根节点

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
typedef struct BtNode* Bt;
struct BtNode
{
    int Num;
    Bt RT;
    Bt LT;
};
vector<vector<int> > Order(2);
Bt BuildTree(Bt T,int i2,int j2,int j1)
{
    if (i2 >= j2)
        return T;
    int i = i2;
    for (; i < j2; i++)
    {
        if (Order[1][i] == Order[0][j1 - 1])
            break;
    }
    if (!T)
    {
        T = new BtNode();
        T->LT = NULL;
        T->RT = NULL;
        T->Num = Order[1][i];
    }
    T->LT =BuildTree(T->LT,i2,i,j1-j2+i);
    T->RT = BuildTree(T->RT,i+1,j2,j1-1);
    return T;
}
int main()
{
    int N;
    cin >> N;
    int num;
    for(int i=0;i<2;i++)
        for (int j = 0; j < N; j++)
        {
            cin >> num;
            Order[i].push_back(num);
        }
    Bt T=NULL;
    T = BuildTree(T,0,N,N);
    queue<Bt> Q;
    Q.push(T);
    cout << T->Num;
    while (!Q.empty())
    {
        if (Q.front()->LT)
        {
            Q.push(Q.front()->LT);
            cout << " " << Q.front()->LT->Num;
        }
        if (Q.front()->RT)
        {
            Q.push(Q.front()->RT);
            cout << " " << Q.front()->RT->Num;
        }
        Q.pop();
    }
    return 0;
}