【PAT】1020 Tree Traversals (25)（25 分）

1020 Tree Traversals (25)（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

 

C++代码如下：

#include<iostream>
#include<queue>
using namespace std;


struct node {
    int data;
    node *lchild, *rchild;
};

int post[35], in[35];
int n;

node* create(int posL,int posR,int inL,int inR) {
    if (posL>posR) return NULL;
    node *root = new node;
    root->data = post[posR];
    int k;
    for (k = inL; k <=inR; k++) {
        if (in[k] == post[posR]) break;
    }
    int numL = k-inL;
    root->lchild = create(posL, posL + numL-1, inL,  k - 1);
    root->rchild = create(posL + numL, posR - 1, k + 1, inR);
    
    return root;
}

int num = 0;

void level(node*r) {
    if (r == NULL) return;
    queue<node*>q;
    q.push(r);
    while (!q.empty()) {
        node *top = q.front();
        q.pop();
        num++;
        cout << top->data;
        if (num < n) cout << ' ';
        if (top->lchild != NULL) q.push(top->lchild);
        if (top->rchild != NULL) q.push(top->rchild);
    }
}

int main() {
    cin >> n;
    int t;
    for (int i = 0; i < n; i++) {
        cin >> t;
        post[i] = t;
    }
    for (int i = 0; i < n; i++) {
        cin >> t;
        in[i] = t;
    }
    node*root = create(0, n - 1, 0, n - 1);
    level(root);
    return 0;
}