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1020. Tree Traversals (序列建树)
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
见之前一篇日志《序列建树（递归） 》
#include <iostream>using namespace std; struct LNode{ LNode *lchild,*rchild; int data;}; int in[31];int pos[31]; LNode* fun(int pos[],int f1,int r1,int in[],int f2,int r2)//生成树{ if(f1>r1) return NULL; LNode *p=(LNode*)malloc(sizeof(LNode)); p->lchild=NULL; p->rchild=NULL; p->data=http://www.mamicode.com/pos[r1]; int i; for(i=f2;i<=r2;i++) if(in[i]==pos[r1]) break; p->lchild=fun(pos,f1,f1+i-f2-1,in,f2,i-1); p->rchild=fun(pos,f1+i-f2,r1-1,in,i+1,r2); return p; } int main(){ int n; while(cin>>n) { int i; for(i=0;i<n;i++) cin>>pos[i]; for(i=0;i<n;i++) cin>>in[i]; LNode *q=(LNode*)malloc(sizeof(LNode)); q=fun(pos,0,n-1,in,0,n-1); LNode* que[40]; //层次遍历 int front=0;int rear=0; rear++; que[rear]=q; bool fir=true; while(front!=rear) { front++; if(fir) {cout<<que[front]->data;fir=false;} else cout<<" "<<que[front]->data; if(que[front]->lchild!=NULL) que[++rear]=que[front]->lchild; if(que[front]->rchild!=NULL) que[++rear]=que[front]->rchild; } cout<<endl; } return 0;}
1020. Tree Traversals (序列建树)