1102. Invert a Binary Tree

最新推荐文章于 2022-03-30 22:47:47 发布

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最新推荐文章于 2022-03-30 22:47:47 发布

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分类专栏： PAT 文章标签： PAT C++

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本文介绍了一种翻转二叉树的算法实现，通过输入节点间的左右子节点关系建立二叉树，并实现了层序输出及右中左遍历输出翻转后的二叉树。示例代码展示了如何构建二叉树并进行遍历。

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

建一棵树，层序输出下，然后右中左遍历一下。

#include<iostream>
#include<vector>
#include<queue>

using namespace std;

typedef struct node
{
	bool f;
	int num;
	node * l;
	node * r;
};

node* p[10];
vector<int>ans;

void bfs(int s)
{
	queue<node*>q;
	node* tmp;
	q.push(p[s]);
	bool f=false;
	while(!q.empty())
	{
		tmp=q.front();
		q.pop();
		if(f)
			cout<<" "<<tmp->num;
		else
		{
			cout<<tmp->num;
			f=true;
		}
		if(tmp->r!=NULL)
			q.push(tmp->r);
		if(tmp->l!=NULL)
			q.push(tmp->l);
	}
}
void dfs(int s)
{
	if(p[s]->r!=NULL)
		dfs(p[s]->r->num);
	ans.push_back(s);
	if(p[s]->l!=NULL)
		dfs(p[s]->l->num);

}
int main()
{
	int n;
	cin>>n;
	for(int i=0;i<n;++i)
	{
		p[i]=new node;
		p[i]->f=false;
		p[i]->num=i;
	}
	for(int i=0;i<n;++i)
	{
		char l,r;
		cin>>l>>r;
		if(l=='-')
			p[i]->l=NULL;
		else
		{
			p[i]->l=p[l-'0'];
			p[l-'0']->f=true;
		}
		if(r=='-')
			p[i]->r=NULL;
		else
		{
			p[i]->r=p[r-'0'];
			p[r-'0']->f=true;
		}
	}
	int root;
	for(int i=0;i<n;++i)
		if(p[i]->f==false)
			root=i;
	bfs(root);
	cout<<endl;
	dfs(root);
	cout<<ans[0];
	for(int i=1;i<ans.size();++i)
		cout<<" "<<ans[i];
	return 0;
}

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