[递归&&bfs]PAT1020 Tree Traversals

1020. Tree Traversals (25)

时间限制

400 ms

内存限制

32000 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

题意:根据题中所给出的后序遍历和中序遍历序列求树的层次遍历。

思路：先建立二叉树，再使用bfs就可以求解。

#include<iostream>#include<queue>using namespace std;int postorder[40],inorder[40];class Node{public:    int data;    Node *ld,*rd;};queue<Node*> st;Node* CreateTree(int n,int *arry1,int *arry2){    if(n<=0) return NULL;    int k=0;    while(arry1[n-1]!=arry2[k])        k++;    Node *root=new Node;    root->data=arry1[n-1];    root->ld=CreateTree(k,arry1,arry2);    root->rd=CreateTree(n-k-1,arry1+k,arry2+k+1);    return root;}int main(){    int n,i,j,k;    cin>>n;    for(i=0;i<n;i++)        cin>>postorder[i];    for(i=0;i<n;i++)        cin>>inorder[i];    Node *root=new Node;    root->ld=NULL;    root->rd=NULL;    root=CreateTree(n,postorder,inorder);    st.push(root);    //cout<<(st.back())->data<<endl;    int tag=0;    while(!st.empty())    {        if((st.front())->ld!=NULL) st.push((st.front())->ld);        if((st.front())->rd!=NULL) st.push((st.front())->rd);        if(tag==0) cout<<(st.front())->data,tag=1;        else cout<<" "<<(st.front())->data;        st.pop();    }    return 0;}