1020. Tree Traversals (25)

来源：互联网发布：pcb开料软件编辑：程序博客网时间：2024/06/02 04:57

1. 原题： https://www.patest.cn/contests/pat-a-practise/1020

2. 思路：

比较简单，先dfs递归利用给出的中序和后序的特点构建唯一的树，然后利用BFS层序输出。

3. 源码（已AC）：

#include<iostream>#include<queue>using namespace std;const int MAX = 30;//最大结点数struct Btree//结点结构体{int data;Btree *left, *right;};int post[MAX] = {0};//存储后序序列int infix[MAX] = {0};//存储中序序列int N;Btree *dfs(int pl, int pr, int il, int ir);//dfs建树void levelOrder(Btree *T);//层序输出void deleteTree(Btree *T);//释放空间，考试时可不写，平时养成好习惯int main(){//freopen("in.txt", "r", stdin);cin >> N;for(int i = 0; i < N; i++)cin >> post[i];for(int i = 0; i < N; i++)cin >> infix[i];Btree *T = dfs(0, N-1, 0, N-1);//cout << 6 << endl;levelOrder(T);return 0;}Btree *dfs(int pl, int pr, int il, int ir)//参数分别表示后序左、边界， 中序左、右边界{if(pl > pr)//递归结束条件return NULL;Btree *t = new Btree;t->data = post[pr];int root;for(int i = 0; i < N; i++)//找出根结点在中序的位置{if(infix[i] == post[pr]){root = i;break;}}t->left = dfs(pl, pr-ir+root-1, il, root-1);//递归左子树， 注意参数，不要写错了，可以在纸上模拟下t->right = dfs(pl+root-il, pr-1, root+1, ir);//递归右子树return t;//返回根结点}void levelOrder(Btree *T)//利用BFS{queue<Btree *> Q;Btree *t = T;Q.push(t);int cnt = 0;//为了便于输出while(!Q.empty()){Btree *tem = Q.front();cnt++;if(tem->left)Q.push(tem->left);if(tem->right)Q.push(tem->right);if(cnt == 1)cout << tem->data;elsecout << " " << tem->data;Q.pop();}cout << endl;}void deleteTree(Btree *T){if(T == NULL)return;deleteTree(T->left);deleteTree(T->right);deleteTree(T);}

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