PAT (Advanced Level) Practise - 1099. Build A Binary Search Tree (30)

 

http://www.patest.cn/contests/pat-a-practise/1099

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
• Both the left and right subtrees must also be binary search trees.

  Given the structure of a binary tree and a sequence of distinct integer keys, there is only one way to fill these keys into the tree so that the resulting tree satisfies the definition of a BST. You are supposed to output the level order traversal sequence of that tree. The sample is illustrated by Figure 1 and 2.

  

  Input Specification:

  Each input file contains one test case. For each case, the first line gives a positive integer N (<=100) which is the total number of nodes in the tree. The next N lines each contains the left and the right children of a node in the format "left_index right_index", provided that the nodes are numbered from 0 to N-1, and 0 is always the root. If one child is missing, then -1 will represent the NULL child pointer. Finally N distinct integer keys are given in the last line.

  Output Specification:

  For each test case, print in one line the level order traversal sequence of that tree. All the numbers must be separated by a space, with no extra space at the end of the line.

  Sample Input:

  9
  1 6
  2 3
  -1 -1
  -1 4
  5 -1
  -1 -1
  7 -1
  -1 8
  -1 -1
  73 45 11 58 82 25 67 38 42
  

  Sample Output:

  58 25 82 11 38 67 45 73 42

 

此题是2015年春季的研究生入学考试复试时的机试题，链接 http://www.patest.cn/contests/graduate-entrance-exam-2015-03-20

这道题的考点有：树的构造、中序遍历（BST，递归）、层序遍历（层序输出，队列）、排序（怎么省事怎么来）。
可能看第一遍时觉得挺复杂的，但是如果树的遍历掌握的话，只要列出处理流程就很简单明了了。

二叉查找树（Binary Search Tree），（又：二叉搜索树，二叉排序树）：
    它或者是一棵空树，或者是具有下列性质的二叉树： 
         若它的左子树不空，则左子树上所有结点的值均小于它的根结点的值； 
         若它的右子树不空，则右子树上所有结点的值均大于它的根结点的值；
         它的左、右子树也分别为二叉排序树。

本题给出节点的左右孩子的下标和一些整型数据，要求构造出一个BST，然后层序输出这颗二叉树上所有节点的key值。

第一步，我们需要构造这棵BST，以便数据归位。实际上，这些节点本身就代表这棵二叉树，这一步使我们在头脑中（概念上）构造二叉树，不需要敲代码实现，考察的对二叉树的理解。
第二步，数据是无序的，所以为了数据归位，需要进行排序。数据量很小，不超过100，所以怎么简单怎么排，反正内存大小和时间足足的.
第三步，进行数据归位。根据bst的性质，其实就是中序处理每个节点。
第四步，按要求层序输出即可。

总结：输入->keys排序->中序遍历，key值归位->层序遍历，格式输出。

#include<cstdio>
#include<cstring> 
// a positive integer N (<=100) 
int num=0,node[100][3]={{0,0,0}},keys[100]={0},keyloc=0;
void levelprint()
{
     int level[100]={0},start=0,len=0;
     level[start]=0,len++;
     while(len)
     {
         if(node[level[start]][1]!=-1)//左子树
             level[start+len]=node[level[start]][1],len++;
         if(node[level[start]][2]!=-1)//左子树
             level[start+len]=node[level[start]][2],len++;
             
         if(start) printf(" ");
         printf("%d",node[level[start]][0]);
         start++,len--;
     }
     
}
void inorder(int rootloc)
{
     if(node[rootloc][1]!=-1)//左子树
         inorder(node[rootloc][1]);
     node[rootloc][0]=keys[keyloc],keyloc++;  //本节点 
    // printf("\n %d  %d  %d  %d",rootloc,node[rootloc][0],node[rootloc][1],node[rootloc][2]);
     if(node[rootloc][2]!=-1)//右子树
         inorder(node[rootloc][2]);
}
void keyssort()
{
     for(int i=1;i<num;i++)
     {
             int key=keys[i],loc=i-1;
             for(;loc>=0;loc--) if(keys[loc]<=key) break;
             loc++;
             for(int j=i;j>loc;j--) keys[j]=keys[j-1];
             keys[loc]=key;
     }
}
int main()
{    
   scanf("%d",&num);
   for(int i=0;i<num;i++) scanf("%d%d",&node[i][1],&node[i][2]);
   for(int i=0;i<num;i++) scanf("%d",keys+i);   
   keyssort();//keys排序 升序
   inorder(0);//中序遍历，keys归位
   levelprint();//层序输出     
   return 0;
}