1102. Invert a Binary Tree (25)

题目链接：http://www.patest.cn/contests/pat-a-practise/1102

题目：

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

分析：

题目中的原意是：给你一个树的输入，让你把数左右互换（invert）然后对其进行层序遍历和中序遍历。

不过可以有一个小trick，对待输入我们不把它当做是先左孩子再右孩子，而是当做是先右孩子再左孩子。

另外，对于这种序号已知的，结构体不用Node* lchild，而是直接用int lchild，会方便很多很多

AC代码：

#include <iostream>
#include<stdio.h>
#include<algorithm>
#include<string>
#include<vector>
#include<queue>
using namespace std;
struct Node{
 int lchild;
 int rchild;
 int parent;
 Node() :lchild(-1), rchild(-1), parent(-1){}
 //因为不是用Node*，所以不能用是否等于NULL来判断
 //所以这里初始化全部用-1，方便用来判断
}buf[10];
int findRoot(int x){//找到根节点
 if (buf[x].parent != -1){
  return findRoot(buf[x].parent);
 }
 return x;
}
vector<int>ans;//为了答案的格式化输出，所以用一个vector存储顺序
void level_order(int root){//层序遍历，需要和queue配合使用
 queue<int>Q;
 Q.push(root);
 while (!Q.empty()){
  int cur = Q.front();
  ans.push_back(cur);
  Q.pop();
  if (buf[cur].lchild != -1)
   Q.push(buf[cur].lchild);
  if (buf[cur].rchild != -1)
   Q.push(buf[cur].rchild);
 }
}
void in_order(int root){//中序遍历，先左再存储再右
 if (buf[root].lchild != -1)
  in_order(buf[root].lchild);
 ans.push_back(root);
 if (buf[root].rchild != -1)
  in_order(buf[root].rchild);
}
void print(vector<int> Vx){//对于vector的格式化输出
 for (int i = 0; i < Vx.size(); ++i){
  if(i) cout << " " << Vx[i];
  else cout << Vx[i];
 }
}
int main(){
 //freopen("F://Temp/input.txt", "r", stdin);
 int n;
 cin >> n;
 for (int i = 0; i < n; i++){
  string l, r;
  cin >> r >> l;//point，先右后左可以直接实现invert树的效果
  if (r != "-"){
   buf[i].rchild = atoi(r.c_str());//孩子和父的两边都要互相记录
   buf[atoi(r.c_str())].parent = i;
  }
  if (l != "-"){
   buf[i].lchild = atoi(l.c_str());
   buf[atoi(l.c_str())].parent = i;
  }
 }
 int root = findRoot(0);//找到根节点
 ans.clear();
 level_order(root);
 print(ans);
 ans.clear();//记得要清空一下
 cout << endl;
 in_order(root);
 print(ans);
 cout << endl;
 return 0;
}

截图：

——Apie陈小旭