PAT 甲级 1020 Tree Traversals (二叉树遍历）

1020. Tree Traversals (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

72 3 1 5 7 6 41 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2

根据后续遍历和中序遍历，求二叉树

#include <iostream>#include <string.h>#include <stdlib.h>#include <algorithm>#include <math.h>#include <stdio.h>#include <queue>using namespace std;typedef struct Tree{    int data;    Tree *lchild;    Tree *rchild;}a[40];int post[40];int in[40];int n;int ans[40];void dfs(int l1,int r1,int l2,int r2,Tree* &root){  root=new Tree();    int i;    for( i=l1;i<=r1;i++)        if(in[i]==post[r2])            break;    root->data=post[r2];    if(i==l1)        root->lchild=NULL;    else        dfs(l1,i-1,l2,l2+i-l1-1,root->lchild);    if(i==r1)        root->rchild=NULL;    else        dfs(i+1,r1,r2-(r1-i),r2-1,root->rchild);}int cnt;void bfs(Tree *tree){  queue<Tree*> q;  q.push(tree);  while(!q.empty())  {    Tree *root=q.front();    q.pop();    ans[cnt++]=root->data;    if(root->lchild!=NULL)      q.push(root->lchild);    if(root->rchild!=NULL)      q.push(root->rchild);  }}int main(){    while(scanf("%d",&n)!=EOF)    {        for(int i=1;i<=n;i++)            scanf("%d",&post[i]);        for(int i=1;i<=n;i++)            scanf("%d",&in[i]);        Tree *tree;        dfs(1,n,1,n,tree);    cnt=0;    bfs(tree);    for(int i=0;i<cnt;i++)    {      if(i==cnt-1)      printf("%d\n",ans[i]);      else        printf("%d ",ans[i]);    }    }    return 0;}