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 = ; i < N; i++) {
         string left, right;
         cin >> left >> right;
         tree[i].left = getNum(right);
         tree[i].right = getNum(left);
         if (tree[i].left != -)
             R[tree[i].left] = false;
         if (tree[i].right != -)
             R[tree[i].right] = false;
     }
     for (int i = ; i < N; i++)
         if (R[i]) {
             root = i;
             break;
         }
     levelOrder(root);
     printf("\n");
     inOrder(root);
     printf("\n");
     return ;
 }
 void inOrder(int t) {
     if (t != -) {
         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 != -) {
             Q.push(tree[t].left);
         }
         if (tree[t].right != -) {
             Q.push(tree[t].right);
         }
         if (!Q.empty())
             printf(" ");
     }
 }
 int getNum(string& s) {
     if (s[] == '-')
         return -;
     int n = ;
     for (int i = ; i < s.length(); i++)
         n = n *  + s[i] - '';
     return n;
 }