1.HDU 1710
题意:已知二叉树的前序遍历和中序遍历,求后序遍历;
思路:把先序序列的第一个数字当作一个根,在中序序列中找到这个数字,则中序序列中根前面的序列为根的左子树,后面的序列为右子树。假如中序历遍中第i个点是根节点,那么划分为两个区域,0~i和i~n-1;
#include<iostream> #include<cstring> using namespace std; int a[1005]; int k; void postorder(int *p, int *q, int len){ if(len == 0)//长度为0,返回 return; int t = 0; int s = *p; for(; t < len; t ++){ if(q[t] == *p){//找到当前中序遍历中第一个与当前前序遍历相同的点 break; } } postorder(p+1, q, t);//求左子树 postorder(p+t+1, q+t+1, len-t-1);//求右子树 a[k++] = s; } int main(){ int n, i, j; int p[1005], q[1005]; while(cin>>n){ for(i = 0; i < n; i ++){ cin>>p[i]; } for(j = 0; j < n; j ++){ cin>>q[j]; } k = 0; postorder(p, q, n); for(i = 0; i < k-1; i ++){ cout<<a[i]<<' '; } cout<<a[k-1]<<endl; } return 0; }
2.拓展:已知中序,后序,求前序;
#include"iostream" using namespace std; int pre[30]; int in[30]; int post[30]; int indexOfRootIn(int start,int stop,int root){ for(int j=start;j<=stop;j++){ if(in[j]==root){ return j; } } return -1; } /*void postOrder(int pre_start,int pre_end,int in_start,int in_end,int length){ if(length==0) return; if(length==1){ cout<<in[in_start]<<endl; return; } int root = pre[pre_start]; int index = indexOfRootIn(in_start,in_end,root); int len1=(index-in_start); int len2=(in_end-index); postOrder(pre_start+1,pre_start+index,in_start,index-1,len1); postOrder(pre_start+index+1,pre_end,index+1,in_end,len2); cout<<root<<" "<<endl; }*/ void preOrder(int post_start,int post_end,int in_start,int in_end,int length){ if(length==0) return; if(length==1){ cout<<in[in_start]<<endl; return; } int root = post[post_end]; int index = indexOfRootIn(in_start,in_end,root); int len1= index-in_start; int len2= in_end-index; cout<<root<<" "<<endl; preOrder(post_start,post_start+index-1,in_start,index-1,len1); preOrder(post_start+index,post_end-1,index+1,in_end,len2); } int main(){ int N; cin>>N; for(int i=0;i<N;i++){ cin>>post[i]; } for(int i=0;i<N;i++){ cin>>in[i]; } preOrder(0,N-1,0,N-1,N); return 0; }