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
#include<cstdio> #include<queue> using namespace std; struct Node{ int data; Node* lchild; Node* rchild; }node; int in[35],post[35],level[35]; int n; Node* create(int postL,int postR,int inL,int inR){ if(postL > postR){ return NULL; } Node* root = new Node; root -> data = post[postR]; int k; for(k = inL; k <= inR; k++){ if(in[k] == post[postR]) break; } int numLeft = k - inL; //左子树的个数 root->lchild = create(postL,postL+numLeft-1,inL,k-1); //左子树个数 root->rchild = create(postL+numLeft,postR-1,k+1,inR); //右子数个数 return root; } int num = 0; void BFS(Node* root){ queue<Node*> q; q.push(root); while(!q.empty()){ Node* now = q.front(); q.pop(); printf("%d",now->data); num++; if(num < n) printf(" "); if(now->lchild != NULL) q.push(now->lchild); //now节点后继 if(now->rchild != NULL) q.push(now->rchild); //2018.8.8 错误1 } } int main(){ scanf("%d",&n); for(int i = 0; i < n; i++){ scanf("%d",&post[i]); } for(int i = 0; i < n; i++){ scanf("%d",&in[i]); } Node* root = create(0,n-1,0,n-1); BFS(root); return 0; }