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 (≤), 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:
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7
Sample Output:
4 1 6 3 5 7 2题意：
后序中序链表建树，求层序
AC代码：
#include<bits/stdc++.h>
using namespace std;
int n;
struct node{
int data;
node *left,*right;
};
int post[];
int in[];
vector<int>level;
queue<node*>q;
node *build(int pl,int pr,int il,int ir){
if(pl>pr || il>ir) return NULL;
int pos=-;
for(int i=il;i<=ir;i++){
if(in[i]==post[pr]){
pos=i;
break;
}
}
node *root=new node();
root->data=post[pr];
root->right=build(pr-(ir-pos),pr-,pos+,ir);//是ir-pos
root->left=build(pl,pr-(ir-pos)-,il,pos-);
return root;
}
int main(){
cin>>n;
for(int i=;i<=n;i++) cin>>post[i];
for(int i=;i<=n;i++) cin>>in[i];
node *root=build(,n,,n);
q.push(root);
while(!q.empty()){
root=q.front();
level.push_back(root->data);
q.pop();
if(root->left) q.push(root->left);
if(root->right) q.push(root->right);
}
for(int i=;i<level.size();i++){
cout<<level.at(i);
if(i!=level.size()-) cout<<" ";
}
return ;
}