PAT(A) 1020

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题意是已知中序遍历和后序遍历，求层序遍历

思路是对于每一棵子树其根，总是在后续遍历的最后，可以递归的求解

一开始没考虑到是BT而不是BST ，左子树并不一定小于右子树，wa了一次

后来deep+1改成左边2*deep，右边2*deep+1，ac

#include<stdio.h>struct list{int value;int deep;} preorder[30];int pp;int compare(struct list a,struct list b){if(a.deep == b.deep)return a.value - b.value;else return a.deep - b.deep;}void PostToPre(int postorder[],int inorder[],int n,int deep){if(n){int povit;povit = postorder[n-1];preorder[pp++].value = povit;preorder[pp-1].deep = deep;int i;for(i = 0;i<n;i++)if(inorder[i] == povit)break;PostToPre(postorder,inorder,i,2*deep+2);PostToPre(postorder+i,inorder+i+1,n-i-1,2*deep+3);}}int main(){int postorder[30],inorder[30];int n,i,j;struct list tmp;scanf("%d",&n);for(i = 0;i<n;i++)scanf("%d",&postorder[i]);for(i = 0;i<n;i++)scanf("%d",&inorder[i]);PostToPre(postorder,inorder,n,0);for(i = 0;i<n;i++)for(j = i+1;j<n;j++){if(compare(preorder[j],preorder[i])<0){tmp = preorder[i];preorder[i] = preorder[j];preorder[j] = tmp;}}for(i = 0;i<n;i++){printf("%d",preorder[i].value);if(i == n-1)printf("\n");else printf(" ");}return 0;}

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