二叉树入门和刷题详解

#include<bits/stdc++.h>

using namespace std;

int a[2000];
int n;

void preorder(int x) //先序
{
	if(x>n) return;
	cout<<x<<" ";
	preorder(2*x);
	preorder(2*x+1);
}

void inorder(int x) //中序
{
	if(x>n) return;
	inorder(2*x);
	cout<<x<<" ";
	inorder(2*x+1);
}

void postorder(int x) //后序 
{
	if(x>n) return;
	postorder(2*x);
	postorder(2*x+1);
	cout<<x<<" ";
}


int main()
{
	cin>>n;
	preorder(1);
	cout<<endl;
	inorder(1);
	cout<<endl;
	postorder(1);
	return 0;
}

# include<bits/stdc++.h>

using namespace std;
typedef pair<int,int> pii;
const int N = 1200;

pii a[N];
int n;

void preorder(int x)
{
	if(x>n || x == 0) return ;
	cout<<x<<" ";
	preorder(a[x].first);
	preorder(a[x].second);
}

void inorder(int x)
{
	if(x>n || x == 0) return ;
	inorder(a[x].first);
	cout<<x<<" ";
	inorder(a[x].second);
}

void postorder(int x)
{
	if(x>n || x == 0) return ;
	postorder(a[x].first);
	postorder(a[x].second);
	cout<<x<<" ";
}


int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		a[i] = {x,y};
	}
	preorder(1);cout<<endl;
	inorder(1);cout<<endl;
	postorder(1);
	return 0;
}

# include<bits/stdc++.h>

using namespace std;

int p[1111];
bool st[1111];

int main()
{
	int n;cin>>n;
	p[1] = 1;
	for(int i=1;i<=n;i++)
	{
		int x,y;cin>>x>>y;
		p[x] = i;p[y] = i;
	}	
	int u,v;cin>>u>>v;
	while(u!=1)
	{
		st[u] = true;u = p[u];   //记录一个点的所有祖先
	}
	st[1] = true;
	while(!st[v]) v = p[v];  //遍历另一个点的所有祖先，第一个和u祖先重合的就是最近公共祖先
	cout<<v<<endl;
	return 0;
}

# include<bits/stdc++.h>

using namespace std;
typedef pair<int,int> pii;

const int N = 1e6+10;

pii a[N];
int ans[N];

int solve(int x)
{
    int t = ans[x];
    if(a[x].first) t+=solve(a[x].first);
    if(a[x].second) t+=solve(a[x].second);
    ans[x] = t;
    return t;
}


int main()
{
    int n;cin>>n;
    for(int i=1;i<=n;i++)
    {
        int x,y;cin>>x>>y;
        a[i] = {x,y};
    }
    for(int i=1;i<=n;i++) {int x;cin>>x;ans[i] = x;}
    solve(1);
    for(int i=1;i<=n;i++) cout<<ans[i]<<" ";
    return 0;
}