PAT 甲级 刷题日记|A 1020 Tree Traversals (25 分)

单词积累

postorder 后序

inorder 中序

level order 层次

题目

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
结尾无空行

Sample Output:

4 1 6 3 5 7 2
结尾无空行

思路

根据中序遍历和后续遍历建树，然后对树进行层次遍历。

关键在于递归进去的边界值。

代码

#include 
using namespace std;

int len;
const int maxn = 1000;
int post[maxn];
int in[maxn];

struct node {
    int data;
    node* leftchild;
    node* rightchild;
    node(int d): data(d), leftchild(NULL), rightchild(NULL) {
    }
};

node* create(int a, int b, int c, int d) {
    if (a > b) {
        return NULL;
    }
    int ro = post[b];
//  cout<leftchild = create(a, a + numleft - 1, c, i - 1);
    root->rightchild = create(a + numleft, b - 1, i + 1, d);
    return root;
}

void level(node *root) {
    queue tra;
    tra.push(root);
    int flag = 0;
    while (!tra.empty()) {
        node* now = tra.front();
        tra.pop();
        if (flag == 0) {
            cout<data;
            flag = 1;
        } else {
            cout<<" "<data;
        }
        if(now->leftchild != NULL) tra.push(now->leftchild);
        if(now->rightchild != NULL) tra.push(now->rightchild);
    }
}

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