problem

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 7Sample Output:4 1 6 3 5 7 2

tip

给出后序与中序遍历，要求层次遍历。

answer

#include
    
     #include
     
      #define Max 33using namespace std;int n, back[Max], mid[Max];//int tree[Max];struct Node{    int d;    Node *l;    Node *r;};Node *root;void Print(){    queue
      
        q;    q.push(root);    while(!q.empty()){        Node *t = q.front(); q.pop();        if(!t->d) continue;        if(!t->l && !t->r && q.empty()) cout<
       
        d;        else cout<
        
         d<<" ";                if(t->l) q.push(t->l);        if(t->r) q.push(t->r);    }//  cout<
         
           right) return NULL; Node *a = new Node(); int num = back[right]; int numIndex = -1; for(int i = midLeft; i <= midRight; i++){ if(mid[i] == num) numIndex = i; } int lNum = numIndex - midLeft, rNum = midRight - numIndex; a->d = num; Print(); a->l = DFS(left, left + lNum -1, numIndex - lNum, numIndex-1); a->r = DFS(right-rNum, right-1, numIndex+1, numIndex+rNum); return a;}int main(){// freopen("test.txt", "r", stdin); ios::sync_with_stdio(false); cin>>n; for(int i = 0; i < n; i++){ cin>>back[i]; } for(int i = 0; i < n; i++){ cin>>mid[i]; } root = new Node(); root = DFS(0, n-1, 0, n-1); Print(); return 0;}
         
        
       
      
     
    

exprience

• 二叉树要多写几次，就熟悉了。