1102 Invert a Binary Tree (25 分) 翻转二叉树并输出层次遍历和中序遍历

深藏阁楼爱情的钟 2024-04-17 23:20 148阅读 0赞

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 (≤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

#include<stdio.h>
    #include<string.h>
    #include<math.h>
    #include<iostream>
    #include<string>
    #include<sstream>
    #include<algorithm>
    #include<map>
    #include<set>
    #include<queue>
    #include<stack>
    #include<vector>
    using namespace std;
    #define inf 0x3f3f3f3f
    #define LL long long
    vector<int> vd;
    int l[20],r[20];
    void dfs(int u)
    {
    	if(r[u]!=-1)dfs(r[u]);
    	vd.push_back(u);
    	if(l[u]!=-1)dfs(l[u]);
    }
    int main()
    {
    	int n,i,j;
    	char x,y;
    	map<int,int> ma;
    	scanf("%d",&n);
    	for(i=0;i<n;i++)
    	{
    		getchar();
    		scanf("%c %c",&x,&y);
    		if(x=='-')
    		l[i]=-1;
    		else
    		l[i]=x-'0',ma[x-'0']++;
    		if(y=='-')
    		r[i]=-1;
    		else
    		r[i]=y-'0',ma[y-'0']++;
    	}
    	int root;
    	for(i=0;i<n;i++)
    	{
    		if(ma[i]==0)
    		root=i;
    	}
    	queue<int> q;
    	vector<int> v;
    	q.push(root);
    	while(!q.empty())
    	{
    		int u=q.front();
    		v.push_back(u);
    		q.pop();
    		if(r[u]!=-1)
    		q.push(r[u]);
    		if(l[u]!=-1)
    		q.push(l[u]);
    	}
    	dfs(root);
    	for(i=0;i<v.size()-1;i++)
    	printf("%d ",v[i]);
    	printf("%d\n",v[v.size()-1]);
    	for(i=0;i<vd.size()-1;i++)
    	printf("%d ",vd[i]);
    	printf("%d\n",vd[vd.size()-1]);
    }