标签：creat 打印 建立 otto sep lock binary ext key
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.
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.
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.
7 2 3 1 5 7 6 4 1 2 3 4 5 6 7
4 1 6 3 5 7 2
1 /* 2 time: 2019-06-30 14:40:45 3 problem: PAT_A1020#Tree Traversals 4 AC: 08:33 5 6 题目大意： 7 给出后序和中序遍历，打印层序遍历 8 */ 9 #include<cstdio> 10 #include<queue> 11 using namespace std; 12 const int M=35; 13 int post[M],in[M],n; 14 struct node 15 { 16 int data; 17 node *lchild,*rchild; 18 }; 19 20 node *Create(int postL, int postR, int inL, int inR) 21 { 22 if(postL > postR) 23 return NULL; 24 node *root = new node; 25 root->data = post[postR]; 26 int k; 27 for(k=inL; k<=inR; k++) 28 if(in[k] == post[postR]) 29 break; 30 int numLeft = k-inL; 31 root->lchild = Create(postL,postL+numLeft-1,inL,k-1); 32 root->rchild = Create(postL+numLeft,postR-1,k+1,inR); 33 return root; 34 } 35 36 void LayerOrder(node *root) 37 { 38 queue<node*> q; 39 q.push(root); 40 int pt=0; 41 while(!q.empty()) 42 { 43 root = q.front(); 44 q.pop(); 45 printf("%d%c", root->data, ++pt==n?‘\n‘:‘ ‘); 46 if(root->lchild) 47 q.push(root->lchild); 48 if(root->rchild) 49 q.push(root->rchild); 50 } 51 } 52 53 int main() 54 { 55 #ifdef ONLINE_JUDGE 56 #else 57 freopen("Test.txt", "r", stdin); 58 #endif // ONLINE_JUDGE 59 60 scanf("%d", &n); 61 for(int i=0; i<n; i++) 62 scanf("%d", &post[i]); 63 for(int i=0; i<n; i++) 64 scanf("%d", &in[i]); 65 node *root = Create(0,n-1,0,n-1); 66 LayerOrder(root); 67 68 return 0; 69 }
标签：creat 打印 建立 otto sep lock binary ext key
原文地址：https://www.cnblogs.com/blue-lin/p/11109732.html
