1020. Tree Traversals (25)

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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：由后序中序遍历求出层序遍历，先构建树，再求层次遍历。

#include <stdio.h>//1#include <iostream>#include <queue>#include <vector>using namespace std;struct Node{Node *l,*r;int num;}node[35];int postseq[35],inseq[35];Node* func(int pl,int pr,int inl,int inr){Node *root=new Node;if(pl>pr) {root=NULL; return root;}for(int i=inl;i<=inr;i++){if(inseq[i]==postseq[pr]){root->num=inseq[i];//cout<<"root->num "<<root->num<<endl;root->l=func(pl,pl+i-inl-1,inl,i-1);root->r=func(pl+i-inl,pr-1,i+1,inr);return root;}}}queue<Node*> q;vector<int> rst;void preout(Node *root){if(root==NULL) return;preout(root->l);preout(root->r);cout<<root->num;}void levelout(Node *root){q.push(root);while(!q.empty()){Node* tmp=q.front();rst.push_back(tmp->num);if(tmp->l) q.push(tmp->l);if(tmp->r) q.push(tmp->r);q.pop();}}int main(){//freopen("E:\input.txt","r",stdin);int n,b;cin>>n;for(int i=0;i<n;i++) cin>>postseq[i];for(int i=0;i<n;i++) cin>>inseq[i];Node *root=func(0,n-1,0,n-1);//cout<<root->num<<endl;//preout(root);levelout(root);for(int i=0;i<rst.size();i++){if(i!=0) cout<<" ";cout<<rst[i];}}

ac代码：

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