1102. Invert a Binary Tree (25)【二叉树】——PAT (Advanced Level) Practise

题目信息

1102. Invert a Binary Tree (25)

时间限制400 ms
内存限制65536 kB
代码长度限制16000 B
The following is from Max Howell @twitter:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.

Now it’s your turn to prove that YOU CAN invert a binary tree!

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=10) which is the total number of nodes in the tree – and hence the nodes are numbered from 0 to N-1. Then N lines follow, each corresponds to a node from 0 to N-1, and gives the indices of the left and right children of the node. If the child does not exist, a “-” will be put at the position. Any pair of children are separated by a space.

Output Specification:

For each test case, print in the first line the level-order, and then in the second line the in-order traversal sequences of the inverted tree. There must be exactly one space between any adjacent numbers, and no extra space at the end of the line.

Sample Input:
8
1 -
- -
0 -
2 7
- -
- -
5 -
4 6
Sample Output:
3 7 2 6 4 0 5 1
6 5 7 4 3 2 0 1

解题思路

二叉树遍历

AC代码

#include <cstdio>
#include <vector>
#include <set>
using namespace std;
vector<int> level[15];
vector<int> inorder;
int get(){
    char s[10];
    scanf("%s", s);
    if (s[0] == ‘-‘) return -1;
    int a;
    sscanf(s, "%d", &a);
    return a;
}
int L[15], R[15];
void inOrder(int root){
    if (R[root] != -1) inOrder(R[root]);
    inorder.push_back(root);
    if (L[root] != -1) inOrder(L[root]);
}
void levelOrder(int root, int lv){
    level[lv].push_back(root);
    if (R[root] != -1) levelOrder(R[root], lv + 1);
    if (L[root] != -1) levelOrder(L[root], lv + 1);
}
int main()
{
    int n;
    scanf("%d", &n);
    set<int> st;
    for (int i = 0; i < n; ++i){
        st.insert(i);
    }
    for (int i = 0; i < n; ++i){
        L[i] = get();
        R[i] = get();
        if (L[i] != -1) st.erase(L[i]);
        if (R[i] != -1) st.erase(R[i]);
    }
    levelOrder(*st.begin(), 0);
    inOrder(*st.begin());
    printf("%d", *st.begin());
    for (int i = 1; i < 15; ++i){
        for (int j = 0; j < level[i].size(); ++j){
            printf(" %d", level[i][j]);
        }
    }
    printf("\n%d", inorder[0]);
    for (int i = 1; i < inorder.size(); ++i){
        printf(" %d", inorder[i]);
    }
    return 0;
}