PAT A1099:Build A Binary Search Tree之中序遍历建树
题目描述
1099 Build A Binary Search Tree (30分)
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
求解思路
首先对输入序列进行预处理排序(升序),利用中序遍历建立一棵二叉搜索树,然后再根据层次遍历序列输出二叉搜索树。
代码实现(AC)
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
struct Node{
int data;
int lchild;
int rchild;
}node[110];
int d[110];
int n;
int index=0;
void inorder(int root)
{
if(root==-1) return;
inorder(node[root].lchild);
node[root].data=d[index++];
inorder(node[root].rchild);
}
void levelorder()
{
queue<int>q;
q.push(0);
int cnt=0;
while(!q.empty())
{
int t=q.front();
q.pop();
if(++cnt!=n) printf("%d ",node[t].data);
else printf("%d",node[t].data);
if(node[t].lchild!=-1) q.push(node[t].lchild);
if(node[t].rchild!=-1) q.push(node[t].rchild);
}
}
void solve()
{
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d %d",&node[i].lchild,&node[i].rchild);
}
for(int i=0;i<n;i++)
{
scanf("%d",&d[i]);
}
sort(d,d+n);
inorder(0);
levelorder();
}
int main()
{
solve();
return 0;
}