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 7Sample Output:
4 1 6 3 5 7 2
代码:
#include <iostream> #include <cstring> #include <queue> using namespace std; struct TNode { int Data; TNode *Left,*Right; }; int h[30],z[30],n; TNode *CreatTNode() { TNode *p = new TNode(); p -> Left = p -> Right = NULL; return p; } TNode *RestoreTree(int z1,int z2,int h1,int h2) { TNode *head = CreatTNode(); head -> Data = h[h2]; for(int i = z2;i >= z1;i --) { if(z[i] == h[h2]) { if(i != z1)head -> Left = RestoreTree(z1,i - 1,h1,h1 + i - z1 - 1); if(i != z2)head -> Right = RestoreTree(i + 1,z2,h1 + i - z1,h2 - 1); break; } } return head; } void levelOrder(TNode *head) { int flag = 0; queue<TNode *>q; q.push(head); while(!q.empty()) { if(q.front() -> Left)q.push(q.front() -> Left); if(q.front() -> Right)q.push(q.front() -> Right); if(flag)cout<<' '<<q.front() -> Data; else { flag = 1; cout<<q.front() -> Data; } q.pop(); } } int main() { cin>>n; for(int i = 0;i < n;i ++) { cin>>h[i]; } for(int i = 0;i < n;i ++) { cin>>z[i]; } TNode *head = RestoreTree(0,n - 1,0,n - 1); levelOrder(head); }