樹的基本構造與遍曆_ACM

題目：1020. Tree Traversals (25)

題目描述：

已知樹的postorder（後序遍曆）與inorder（中序遍曆），構造出完整的樹，並按層次遍曆輸出數

代碼展示一（已知後與中，進行層次遍曆）：

#include <iostream>#include <queue>using namespace std;const int maxn=1e3+5;int a[maxn],b[maxn];struct Tree{Tree *left,*right;int data;};Tree *buildtree(int la,int ra,int lb,int rb){if (la>ra) return NULL;int p;for (int i=lb;i<=rb;i++) if (b[i]==a[ra]) {p=i;break;}Tree *tree=new Tree;tree->data  = a[ra];tree->left  = buildtree (la,la+(p-lb)-1,lb,p-1);tree->right = buildtree (la+(p-lb),ra-1,p+1,rb);return tree;}void print_levelorder(Tree *root){queue < Tree * > q;cout<<root->data;if (root->left!=NULL)  q.push(root->left);if (root->right!=NULL) q.push(root->right);while(!q.empty()){Tree *tree=new Tree;tree=q.front();q.pop();cout<<" "<<tree->data;if (tree->left!=NULL)  q.push(tree->left);if (tree->right!=NULL) q.push(tree->right);}}int main(){//freopen("datain.txt","r",stdin);int n;cin>>n;for (int i=1;i<=n;i++) cin>>a[i];for (int i=1;i<=n;i++) cin>>b[i];Tree *root=buildtree(1,n,1,n);print_levelorder(root);}

代碼展示二（已知前與中，進行層次遍曆）：

只需更改建樹的部分代碼即可

Tree *buildtree(int la,int ra,int lb,int rb){if (la>ra) return NULL;int p;for (int i=lb;i<=rb;i++) if (b[i]==a[la]) {p=i;break;}Tree *tree=new Tree;tree->data  = a[la];tree->left  = buildtree (la+1,la+(p-lb),lb,p-1);tree->right = buildtree (la+(p-lb)+1,ra,p+1,rb);return tree;}