PAT 1119. Pre- and Post-order Traversals (30) 用前序、后序求中序
2017-09-16 10:25
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// VSPAT.cpp : 定义控制台应用程序的入口点。 // #include<cmath> #include<stdio.h> #include<algorithm> #include<cstring> #include<iostream> #include<stack> #include<vector> #include<queue> #include<string> #include<map> using namespace std; //70min //耗时于思考和改bug /************************* 题意:给出前序和后序,求中序 若中序不唯一,输出任意一个,并输出No *************************/ /************************ 求解要点: 对于一棵树 前序序列的首位和后序序列的末位一定相同,且为根 前序根后的那位一定为“左”子树根。 后序根前的那位一定为“右”子树。 前: 根 左子树根 XXXXXXX 后: XXXXXXX 右子树根 根 然后在后序序列中找到左子树根,左子树根的右边都是左子树 前: 根 [左子树根 XXX][右子树根XXX] 后: [XXX左子树根] [XXX右子根] 根 根据如此划分进行递归 ★若左树根=右树根,则仅1个子树,故极可能是左也可能是右 这里同意划分为左根,右孩子=-1 ************************/ /*********************** 笔记: 注意输入中,树的数字可能不为1~N,可能为很大的整数。 故用数组做树结构时要留意大小。 *********************/ #define M 1000 int pre[M],post[M]; int Left[M],Right[M]; int flag = 1,n; int buildtree(int node,int prel,int prer,int postl,int postr) { int i,lchild,rchild; if(prel==prer||prer>=n||prel<0||postl<0||postr>=n) { Left[node]=-1; Right[node]=-1; return node; } //printf("nownode=%d\n",node); lchild=pre[prel+1]; rchild=post[postr-1]; if(lchild==rchild) { flag=0; Left[node]=buildtree(lchild,prel+1,prer,postl,postr-1); Right[node]=-1; //printf("node=%d,left=%d,right=%d\n",node,Left[node],Right[node]); return node; } for(i=postr-1;i>=postl;i--) if(post[i]==lchild) break; //左子树为 前序左到中, 后序左到中 //右子树为 前序中到右 后序中到右 Left[node]=buildtree(lchild,prel+1,prel+1+i-postl,postl,i); Right[node]=buildtree(rchild,prel+i-postl+2,prer,i+1,postr-1); return node; } int k=0; void orderprint(int node) { if(node<=0) return ; orderprint(Left[node]); printf("%d",node); k++; if(k<n) cout<<" "; else cout<<endl; orderprint(Right[node]); } int main() { int i; scanf("%d",&n); for(i=0;i<n;i++) scanf("%d",&pre[i]); for(i=0;i<n;i++) scanf("%d",&post[i]); int root=pre[0]; buildtree(root,0,n-1,0,n-1); if(flag==1) cout<<"Yes"<<endl; else cout<<"No"<<endl; orderprint(root); return 0; }
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