PAT L2 011 玩转树遍历 建树加BFS

通过两个已知遍历结果可以快速通过二分建二叉树，而层序遍历只要Bfs时把每个按顺序放入即可；

#include <stdio.h>  
#include <algorithm>  
#include <string.h>  
#include <queue>  
#include <iostream>  
  
using namespace std;  
  
typedef long long ll;  
const int N = 35;  
  
int cnt, n;  
  
typedef struct Tree  
{  
    Tree *left;  
    Tree *right;  
    int val;  
}Tree;  
  
Tree *root;  
  
Tree* creat(int *preorder, int *inorder, int len)
{  
    Tree *tmp;  
    for(int i = 0; i < len; i++)  
    {  
        if(preorder[0]==inorder[i])
        {  
            tmp = (Tree*)malloc(sizeof(Tree));  
            tmp->val = inorder[i];  
            tmp->left = creat(preorder+1, inorder, i);  
            tmp->right = creat(preorder+i+1, inorder+i+1,len-(i+1));  
            return tmp;  
        }  
    }  
    return NULL;  
}  
  
void levelorder(Tree *cur)  
{  
    cnt = 0;  
    queue<Tree*>que;  
    while(!que.empty()) que.pop();  
    que.push(cur);  
    Tree *node;  
    while(!que.empty())  
    {  
        node = que.front();  
        que.pop();  
        if(cnt == n-1) printf("%d\n", node->val);  
        else  
        {  
            printf("%d ", node->val);  
            cnt++;  
        }  
        if(node->right != NULL) que.push(node->right);  
        if(node->left != NULL) que.push(node->left);  
    }  
}  
  
int main()  
{  
  //  freopen("in.txt", "r", stdin);  
    int preorder[N], inorder[N];  
    while(~scanf("%d", &n))  
    {  
        for(int i = 0; i < n; i++)  
            scanf("%d", &inorder[i]);  
        for(int i = 0; i < n; i++)  
            scanf("%d", &preorder[i]);  
        root = creat(preorder, inorder, n);  
        levelorder(root);  
    }  
}