二叉树的几种存储方式（一个数组，指针形式，两个数组）

用一个数组存储完全二叉树 若一个根节点索引为k，则其左子树索引为2k,右子树索引为2k+1;
见下一题 题目来自UVA６７９

# include
# include
using namespace std;
/*
input:树的深度 D , the number of balls:i
output:最后一个小球的位置；
节点为0，走左 且置1 ； 
节点为1，走右 且置0； 
*/
const int max_n=20;
int tree[1<>m;
	while(m--)
	{
		cin>>d;
		if(d==-1) return 0;
		cin>>i;
		long long dd=(1<dd)
				     break;
		    }
			
		}
		cout<

指针形式的二叉树
见下题ｕｖａ１２２

# include
# include 
# include
# include
using namespace std;
const int maxn=100000;
struct node{
	int v;
 	bool have_value;
	node*left,*right;
	node()
	{
		have_value=false;
		left=NULL;
		right=NULL;
		}	
};
node *root;
vector ans;
bool failed;
char s[maxn];
node* newnode()
{
	return new node();
}
void addnode(int ,char*);
bool read_input()
{
	failed=false;
	root=newnode();
	for(;;)
	{
		if(scanf("%s",s)!=1) return false;
		if(!strcmp(s,"()")) break;
		int v;
		sscanf(&s[1],"%d",&v);
		addnode(v,strchr(s,',')+1); 
	}
	return true;
}
void addnode(int v,char*s)
{
	int n=strlen(s);
	node *u=root;
	for(int i=0;ileft==NULL) u->left=newnode();
			u=u->left;
			
		}
		else if(s[i]=='R')
		{
			if(u->right==NULL) u->right=newnode();
			u=u->right;
		}	
	 } 
	 if(u->have_value) failed=true;
	 u->v=v;
	 u->have_value=true;
}
bool bfs(vector&ans)
{
	queue q;
	ans.clear();
	q.push(root);
	while(!q.empty())
	{
		node *u=q.front();q.pop();
		if(!u->have_value)  return false;
		ans.push_back(u->v);
		if(u->left) q.push(u->left);
		if(u->right) q.push(u->right);
	}
	return true;
}
int main()
{
	read_input();
	failed=bfs(ans);
	if(!failed) {
	cout<<"-1";
	 return 0;
     } 
	for(vector ::iterator in=ans.begin();in!=ans.end();++in)
	{
		cout<<*in<<" ";
	}
	return 0;
}

用两个数组存放二叉树
见ｕｖａ５４８
根节点ｒｏｏｔ　l_tree[root]为左子树，r_tree[root]为右子树

# include
# include
# include
using namespace std;
//通过中序 和后序 输出先序
const int max_n=100000;
int in_order[max_n];
int post_order[max_n];
int r_tree[max_n];
int l_tree[max_n];
int n;
bool readlist(int *a)
{
	string line;
	n=0;   //即调用一次readlist()需赋值。因为不重复赋值n的值会一直加下去； 
	int x;
	if(!getline(cin,line)) return false;
	stringstream ss(line);
	while(ss>>x) a[n++]=x;
	return n>0;
 }
 int build(int l1,int r1,int l2,int r2)
 {
 	if(l1>r1) return false;
 	int root=post_order[r2];
 	int p=l1;
 	while(in_order[p]!=root) p++;
 	int cnt=p-l1;
 	l_tree[root]=build(l1,p-1,l2,l2+cnt-1);
 	r_tree[root]=build(p+1,r1,l2+cnt,r2-1);
 	return root;
  } 
  int best,best_sum;
  void dfs(int u,int sum)
  {
  	sum+=u;
  	if(!l_tree[u]&&!r_tree[u])
  	{
	  if(sum