solution Of Pat 1099. Build A Binary Search Tree (30)

1099. Build A Binary Search Tree (30)

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.



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

结题思路 :

题意要求我们将数的序列放入到二叉树中，并进行层次输出。

要求1：保证序列只会有一种在二叉树中的存在方式；

要求2：层次输出树形图即可；

程序步骤：

第一步：对节点的索引进行中序遍历，即此时的对应的数字的排序理应是从小到大的。

第二步：将数字从小到大排序，放入对应的索引的位置。

第三步：层次输出。

具体程序(AC)如下：

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
struct node
{
int value;
int left;
int right;
};
vector<node> tree;
vector<int> list;//暂存索引
vector<int> value;//暂存真实数值
vector<int> result;//保存层次遍历的结果
queue<int> Q;//层次遍历借用的队列
void LDR(int root)//中序遍历
{
if(root==-1)
return;
LDR(tree[root].left);
list.push_back(root);
LDR(tree[root].right);
}
void level(int root)//层次遍历
{
Q.push(root);
int cur;
while(!Q.empty())
{
cur=Q.front();
result.push_back(tree[cur].value);
Q.pop();
if(tree[cur].left!=-1)
Q.push(tree[cur].left);
if(tree[cur].right!=-1)
Q.push(tree[cur].right);

}
}
int ROOT=0;
int main() {
// your code goes here
int n;
cin>>n;
if(n==0)
return 0;
node tmp;
tree.clear();
list.clear();
value.clear();
result.clear();
while(!Q.empty())
Q.pop();
for(int i=0;i<n;++i)
{
cin>>tmp.left>>tmp.right;
tree.push_back(tmp);
}

LDR(ROOT);
int nodeV;
for(int i=0;i<n;++i)
{
cin>>nodeV;
value.push_back(nodeV);
}
sort(value.begin(),value.end());
int loc;
for(int i=0;i<n;++i)
{
loc=list[i];
tree[loc].value=value[i];
}
level(ROOT);
cout<<result[0];
for(int i=1;i<result.size();++i)
cout<<" "<<result[i];
cout<<endl;
return 0;
}