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
判题程序
Standard
作者
CHEN, Yue
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
输出反转二叉树的层次和先序遍历。
#include<cstdio>
#include<vector>
#include<queue>
#include<string>
#include<map>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int INF = 0x7FFFFFFF;
const int maxn = 1e5 + 10;
int n, ch[maxn][2], flag[maxn];
char s[maxn];
int get()
{
scanf("%s", s);
return s[0] == '-' ? -1 : s[0] - '0';
}
void dfs(int x, int &flag)
{
if (x == -1) return;
dfs(ch[x][1], flag);
if (flag) printf(" "); else flag = 1;
printf("%d", x);
dfs(ch[x][0], flag);
}
int main()
{
scanf("%d", &n);
for (int i = 0; i < n; i++)
{
ch[i][0] = get(); if (ch[i][0] != -1) flag[ch[i][0]] = 1;
ch[i][1] = get(); if (ch[i][1] != -1) flag[ch[i][1]] = 1;
}
for (int i = 0; i < n; i++)
{
if (flag[i]) continue;
queue<int> p; p.push(i);
while (!p.empty())
{
int q = p.front(); p.pop();
printf("%s%d", i == q ? "" : " ", q);
if (ch[q][1] != -1) p.push(ch[q][1]);
if (ch[q][0] != -1) p.push(ch[q][0]);
}
printf("\n");
int x = 0; dfs(i, x);
printf("\n");
}
return 0;
}