树的几种遍历

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
string in, level; 

void preorder(int il, int ir, int ll, int lr)
{
    int pos;
    for (int i = ll; i <= lr; i++)
    {
        int flag = 0;
        for (int j = il; j <= ir; j++)//找根结点
        {
            if (level[i] == in[j])
            {
                cout << in[j];
                pos = j;
                flag = 1;
                break;
            }
        }
        if (flag) break;
    }
    if (pos > il)preorder(il, pos - 1, ll, lr);// 遍历左子树
    if (pos < ir) preorder(pos + 1, ir, ll, lr);// 遍历右子树
}


int main()
{
    cin >> in >> level; 
    preorder(0, in.size() - 1, 0, level.size() - 1);//中序区间
    return 0;
}

acwing y总代码

#include<iostream>
#include<cstring>
#include<string>
#include<unordered_map>
#include<queue>

using namespace std;

const int N = 40;
int n;
int postorder[N], inorder[N];
unordered_map <int, int> l, r, pos;//左儿子，右儿子，中序遍历中这一点的下标

int build(int il, int ir, int pl, int pr)//建树，返回的是根节点
{
	int root = postorder[pr];
	int k=pos[root];//k是中序遍历的下标
	if (il < k)l[root] = build(il, k - 1, pl, k - 1 - il + pl);
	if (ir > k)r[root] = build(k + 1, ir, k - 1 - il + pl + 1, pr - 1);
	return root;
}

void bfs(int root)
{
	queue<int>q;
	q.push(root);
	while (q.size())
	{
		auto t = q.front();
		q.pop();
		cout << t << ' ';
		if (l.count(t))q.push(l[t]);//左子树存在
		if (r.count(t))q.push(r[t]);//右子树存在
	}
}
int main()
{
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> postorder[i];
	for (int i = 0; i < n; i++)
	{
		cin >> inorder[i];
		pos[inorder[i]] = i;
	}

	int root = build(0, n - 1, 0, n - 1);//当前结点的中序遍历区间和后序遍历区间
	bfs(root);
	return 0;
}

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;

const int N = 40;
int l[N],r[N];  //l[i],r[i]代表i号节点的左右儿子
int post[N], pre[N]; //后续遍历和前序遍历的序列
bool is_only = true;  //全局变量，是否为唯一二叉树的标志
int in[N], cnt = 0;

//根据pre[l1,r1], post[l2,r2]创建二叉树节点，并分配左右儿子指针
int build(int l1, int r1, int l2, int r2)
{
    int root = pre[l1];

    if(l1 == r1) return root;
    //根据pre[l1+1] 和 post[r2-1]来判断答案是否为1
    if(pre[l1+1] == post[r2-1]) //说明答案不唯一，我们直接认为右儿子不存在即可
        l[root] = build(l1+1, r1, l2, r2-1), is_only = false;
    else  //不相同，那么说明 pre[l1+1]为左儿子的根节点， post[r2-1]为右儿子根节点
    {
        int lpre, rpost; //lpre代表前序序列中左子树的右边界，rpost代表后序序列中右子树的左边界
        for(lpre = l1+1;lpre<=r1;lpre++) if(pre[lpre] == post[r2-1]) break;
        for(rpost = r2-1;rpost>=l2;rpost--) if(post[rpost] == pre[l1+1]) break;
        lpre--, rpost++;
        l[root] = build(l1+1, lpre, l2, rpost-1);
        r[root] = build(lpre+1, r1, rpost, r2-1);
    }

    return root;
}

//中序遍历
void inorder(int root)
{
    if(l[root] != -1) inorder(l[root]);
    in[cnt++] = root;
    if(r[root] != -1) inorder(r[root]);
}

int main()
{
    //用-1来来代表儿子节点为空
    memset(l,-1,sizeof(l));
    memset(r,-1,sizeof(r));
    int n;
    cin >> n;
    for(int i=0;i<n;i++)
        cin >> pre[i];
    for(int i=0;i<n;i++)
        cin >> post[i];

    //根据前序遍历和后序遍历创建二叉树
    int root = build(0,n-1,0,n-1);

    if(is_only) puts("Yes");
    else puts("No");
    inorder(root);

    cout << in[0];
    for(int i=1;i<n;i++)
        cout << " " <<in[i];
}

作者：SmallK
链接：https://www.acwing.com/solution/content/14292/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

https://vjudge.net/contest/492108#problem/A

大佬原文：https://blog.youkuaiyun.com/laojiu_/article/details/51527152
我只是保存一下

#include<iostream>    
#include<algorithm>    

using namespace std;


//前序和中序得到后序

struct Node
{
	int data;
	Node* pLeft;
	Node* pRight;
	Node(int _data):data(_data),pLeft(NULL),pRight(NULL)
	{}
};

Node* root;
int N;
int a[1005];//前序
int b[1005];//中序

Node* create(int t1, int t2, int num)
{
	Node* p = nullptr;

	for (int i = 0; i < num; i++)
	{
		if (a[t1] == b[t2 + i])
		{
			p = new Node(a[t1]);
			p->pLeft = create(t1 + 1, t2, i);
			p->pRight = create(t1 + i + 1, t2 + i + 1, num - i - 1);//num - i - 1下标从零开始，所以减一
			return p;
		}
	}
	return p;
}

void postOrder(Node* p)
{
	if (p == nullptr)
		return;
	else
	{
		postOrder(p->pLeft);
		postOrder(p->pRight);
		if (p == root)
			printf("%d\n", p->data);
		else
			printf("%d ", p->data);
	}
}

int main()
{
	while (~scanf("%d", &N))
	{
		for (int i = 0; i < N; i++)
			scanf("%d", &a[i]);//前序
		for (int i = 0; i < N; i++)
			scanf("%d", &b[i]);//中序
		root = create(0, 0, N);//前序区间，个数
		postOrder(root);
	}
	return 0;
}