A1020

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 (≤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
这道题简直就是经典中的经典，是数据结构的经典实现，英文内容的意思就是，已知后序与中序遍历，求层次遍历，
具体实现方法分两步走：
　　①：由后序与中序的遍历构造树，这在create函数中实现，树的构造用迭代来实现。
　　②：层次遍历对一个树中的所有结点进行遍历。
代码如下：
#include <cstdio>
#include <queue>
#include <cstring>
const int maxn = 50;
using namespace std;
int n;
int pre[maxn], in[maxn], post[maxn];
struct node {
    int data;
    node* lchild;
    node* rchild;
};
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 numLeft  = k - inL;
    root ->lchild = create(postL, postL + numLeft - 1, inL, k - 1);
    root ->rchild = create(postL + numLeft, postR - 1, k + 1, inR);
    return root;
}
int num = 0;
void BFS(node* root){
    queue<node*> q;
    q.push(root);
    while(!q.empty()){
        node* now = q.front();
        q.pop();
        printf("%d", now -> data);
        num++;
        if(num < n) printf(" ");
        if(now ->lchild != NULL)    q.push(now ->lchild);
        if(now ->rchild != NULL)    q.push(now ->rchild);
    }
}
int main() {
    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);
    return 0;
}
其中有一些变量只是说明型的，没有具体应用上。