PAT甲级1099
原题:
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
题目大意:给出总结点数n,然后给出第i个节点的左右孩子,根从0开始,若是-1则到底了。求出层序序列
由于题目说的是BST,所以给出序列从小到大排序即为中序序列,可以通过中序建树,然后再进行层序遍历即可,因为根始终从0开始所以不用额外求根
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
vector<int> in,level;
struct node{
int data; //节点存放的数据
int left,right; //左孩子右孩子
};
node nd[101];
int cnt=0;
void inOrder(int root){
if(root==-1) return ;
inOrder(nd[root].left);
nd[root].data=in[cnt++];
inOrder(nd[root].right);
}
void levelOrder(int root){
if(root==-1) return ;
queue<int> q;
q.push(root);
while(!q.empty()){
int tmp=q.front();
q.pop();
level.push_back(tmp);
if(nd[tmp].left!=-1) q.push(nd[tmp].left);
if(nd[tmp].right!=-1) q.push(nd[tmp].right);
}
}
int main(){
int n,tmpleft,tmpright,tmp;
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%d%d",&tmpleft,&tmpright);
nd[i].left=tmpleft;
nd[i].right=tmpright;
}
for(int i=0;i<n;i++){
scanf("%d",&tmp);
in.push_back(tmp);
}
sort(in.begin(),in.end());
inOrder(0);
levelOrder(0);
for(int i=0;i<level.size();i++){
if(i!=0) printf(" ");
printf("%d",nd[level[i]].data);
}
system("pause");
return 0;
}