1020. Tree Traversals (25)

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

---

代码：

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
struct TNode
{
    int Data;
    TNode *Left,*Right;
};
int h[30],z[30],n;
TNode *CreatTNode()
{
    TNode *p = new TNode();
    p -> Left = p -> Right = NULL;
    return p;
}
TNode *RestoreTree(int z1,int z2,int h1,int h2)
{
    TNode *head = CreatTNode();
    head -> Data = h[h2];
    for(int i = z2;i >= z1;i --)
    {
        if(z[i] == h[h2])
        {
            if(i != z1)head -> Left = RestoreTree(z1,i - 1,h1,h1 + i - z1 - 1);
            if(i != z2)head -> Right = RestoreTree(i + 1,z2,h1 + i - z1,h2 - 1);
            break;
        }
    }
    return head;
}
void levelOrder(TNode *head)
{
    int flag = 0;
    queue<TNode *>q;
    q.push(head);
    while(!q.empty())
    {
        if(q.front() -> Left)q.push(q.front() -> Left);
        if(q.front() -> Right)q.push(q.front() -> Right);
        if(flag)cout<<' '<<q.front() -> Data;
        else
        {
            flag = 1;
            cout<<q.front() -> Data;
        }
        q.pop();
    }
}
int main()
{
    cin>>n;
    for(int i = 0;i < n;i ++)
    {
        cin>>h[i];
    }
    for(int i = 0;i < n;i ++)
    {
        cin>>z[i];
    }
    TNode *head = RestoreTree(0,n - 1,0,n - 1);
    levelOrder(head);
}