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: 7 2 3 1 5 7 6 4 1 2 3 4 5 6 7 Sample Output: 4 1 6 3 5 7 2 模板题，但是还是不够熟练， post 后序 in 中序 pre 前序 level ,layer 层次 还要强调一点层次遍历时，queue<node*>队列存的是地址，还像是queue会创建一个副本什么的，和insert 树（往树里面插入一个节点），理解但还似乎没有透彻，先记住吧
#include<iostream> #include<cstdio> #include<algorithm> #include<queue> const int maxn = 35; using namespace std; struct node{ int date; node* lchild; node* rchild; }; int post[maxn],in[maxn],n; node* creat(int postl,int postr,int inl,int inr){ if(postl > postr){ return NULL; } node* root = new node; root->date = post[postr]; int k; for(k=inl;k<=inr;k++){ if(in[k] == post[postr]) break; } int num=k-inl; root->lchild=creat(postl,postl+num-1,inl,k-1); root->rchild=creat(postl+num,postr-1,k+1,inr); return root; } int num=0; void bfs(node* root){ queue<node*>q; q.push(root); while(!q.empty()){ node* top = q.front(); q.pop(); num++; if(num < n){ cout<<top->date<<' '; } else cout<<top->date; if(top->lchild != NULL) q.push(top->lchild); if(top->rchild != NULL) q.push(top->rchild); } } int main(){ cin>>n; for(int i=0;i<n;i++) cin>>post[i]; for(int i=0;i<n;i++) cin>>in[i]; node* root = creat(0,n-1,0,n-1); bfs(root); return 0; }