hdu3999The order of a Tree (二叉平衡树（AVL）)
Problem Description
As we know,the shape of a binary search tree is greatly related to the order of keys we insert. To be precisely:
1. insert a key k to a empty tree, then the tree become a tree with
only one node;
2. insert a key k to a nonempty tree, if k is less than the root ,insert
it to the left sub-tree;else insert k to the right sub-tree.
We call the order of keys we insert “the order of a tree”,your task is,given a oder of a tree, find the order of a tree with the least lexicographic order that generate the same tree.Two trees are the same if and only if they have the same shape.
1. insert a key k to a empty tree, then the tree become a tree with
only one node;
2. insert a key k to a nonempty tree, if k is less than the root ,insert
it to the left sub-tree;else insert k to the right sub-tree.
We call the order of keys we insert “the order of a tree”,your task is,given a oder of a tree, find the order of a tree with the least lexicographic order that generate the same tree.Two trees are the same if and only if they have the same shape.
Input
There are multiple test cases in an input file. The first line of each testcase is an integer n(n <= 100,000),represent the number of nodes.The second line has n intergers,k1 to kn,represent the order of a tree.To make if more simple, k1 to kn is a sequence of 1 to n.
Output
One line with n intergers, which are the order of a tree that generate the same tree with the least lexicographic.
Sample Input
4 1 3 4 2
Sample Output
1 3 2 4
Source
#include<stdio.h>
#include<malloc.h>
typedef struct tree
{
struct tree *lchilde,*rchilde;
int w;
}tree,*Tree;
void creaRoot(Tree &root)//创建根节点
{
root=(Tree)malloc(sizeof(tree));
root->lchilde=root->rchilde=NULL;
}
void printTree(Tree root,int k)//遍历树并输出该节点的值
{
if(root==NULL) return ;//为空时反回
if(k==0)
printf("%d",root->w);
else
printf(" %d",root->w);
printTree(root->lchilde,k+1);//先搜左
printTree(root->rchilde,k+1);//后搜右
}
void InTree(Tree &node,int x)//建立二叉树，比当前节点小的放左边，否则放右边
{
Tree T;
if(node==NULL)//为空时，就把该值放到该节点
{
T=(Tree)malloc(sizeof (tree));
T->w=x; T->lchilde=T->rchilde=NULL;//把左右孩子置为空
node=T;
}
else//不为空，继续搜
{
if(x<=node->w)
InTree(node->lchilde,x);
else
InTree(node->rchilde,x);
}
}
int main()
{
Tree root;
int n,i,x;
while(scanf("%d",&n)>0)
{
if(n==0)continue;
creaRoot(root);
scanf("%d",&x);
root->w=x;
for(i=2;i<=n;i++)
{
scanf("%d",&x);
InTree(root,x);
}
printTree(root,0);
printf("\n");
}
}