PAT甲级——1102 Invert a Binary Tree (层序遍历+中序遍历)

本文同步发布在CSDN：https://blog.csdn.net/weixin_44385565/article/details/90577042

1102 Invert a Binary Tree (25 分)

 

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 (≤) 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

题目大意：将二叉树每个节点的左右孩子交换位置，然后分别层序和中序输出。第 i 行代表了节点 i 的左右孩子的信息，先左后右，'-' 表示没有该方向的子节点。

思路：用数组tree存储树。

读取字符串然后将其转成int型变量，空节点'-' 用-1代替。

只要在读取数据的时候交换左右孩子的位置就行。

读取数据的时候用bool数组R对每个孩子节点进行标记，然后遍历数组寻找根节点（即没有被标记过的节点）

#include <iostream>
#include <vector>
#include <string>
#include <queue>
using namespace std;
struct node {
    int left, right;
};
vector <node> tree;
bool flag = false;//用于中序遍历标记第一个输出的节点
int getNum(string &s);
void levelOrder(int t);
void inOrder(int t);
int main()
{
    int N, root;
    scanf("%d", &N);
    tree.resize(N);
    vector <bool> R(N, true);
    for (int i = 0; i < N; i++) {
        string left, right;
        cin >> left >> right;
        tree[i].left = getNum(right);
        tree[i].right = getNum(left);
        if (tree[i].left != -1)
            R[tree[i].left] = false;
        if (tree[i].right != -1)
            R[tree[i].right] = false;
    }
    for (int i = 0; i < N; i++)
        if (R[i]) {
            root = i;
            break;
        }
    levelOrder(root);
    printf("
");
    inOrder(root);
    printf("
");
    return 0;
}
void inOrder(int t) {
    if (t != -1) {
        inOrder(tree[t].left);
        if (flag)
            printf(" ");
        if (!flag)
            flag = true;
        printf("%d", t);
        inOrder(tree[t].right);
    }
}
void levelOrder(int t) {
    queue <int> Q;
    Q.push(t);
    while (!Q.empty()) {
        t = Q.front();
        printf("%d", t);
        Q.pop();
        if (tree[t].left != -1) {
            Q.push(tree[t].left);
        }
        if (tree[t].right != -1) {
            Q.push(tree[t].right);
        }
        if (!Q.empty())
            printf(" ");
    }
}
int getNum(string& s) {
    if (s[0] == '-')
        return -1;
    int n = 0;
    for (int i = 0; i < s.length(); i++)
        n = n * 10 + s[i] - '0';
    return n;
}