PAT_A1099#Build A Binary Search Tree

Source:

PAT A1099 Build A Binary Search Tree (30 分)

Description:

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 (≤) 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 − 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

Keys:

• 二叉树的建立
• 二叉树的遍历
• 二叉查找树（Binary Search Tree）

Code:

 /*
 Data: 2019-06-26 16:22:18
 Problem: PAT_A1099#Build A Binary Search Tree
 AC: 21:23

 题目大意：
 BST定义：lchild < root <= rchild
 给定一棵BST的树形，将所给序列填入BST中，并打印层次遍历
 输入：
 第一行给出，结点数N<=100
 接下来N行，给出第i号结点，左孩子的序号，右孩子的序号（0~N-1，-1为空，0为根）
 最后一行，N个数
 输出：
 层次遍历

 基本思路：
 建立静态树，构建根结点与左右孩子之间的关系
 中序遍历静态树，由于BST的中序遍历就是从小到大递增的序列，把排序好的序列依次填入树的结点中
 层次遍历并打印输出
 */
 #include<cstdio>
 #include<queue>
 #include<vector>
 #include<algorithm>
 using namespace std;
 const int M=;
 priority_queue<int,vector<int>,greater<int> > bst;
 struct node
 {
     int data;
     int lchild,rchild;
 }tree[M];

 void InOrder(int root)
 {
     if(root == -)
         return;
     InOrder(tree[root].lchild);
     tree[root].data = bst.top();
     bst.pop();
     InOrder(tree[root].rchild);
 }

 void Travel(int root, int len)
 {
     queue<int> q;
     q.push(root);
     while(!q.empty())
     {
         root = q.front();
         q.pop();
         printf("%d%c", tree[root].data, --len==?'\n':' ');
         if(tree[root].lchild != -)
             q.push(tree[root].lchild);
         if(tree[root].rchild != -)
             q.push(tree[root].rchild);
     }
 }

 int main()
 {
 #ifdef ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif // ONLINE_JUDGE

     int n,x;
     scanf("%d", &n);
     for(int i=; i<n; i++)
         scanf("%d %d", &tree[i].lchild,&tree[i].rchild);
     for(int i=; i<n; i++)
     {
         scanf("%d", &x);
         bst.push(x);
     }
     InOrder();
     Travel(,n);

     return ;
 }