1099. Build A Binary Search Tree (30)

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

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

分析：

思路：首先把二叉排序树中的每个节点的左右子树的数目给统计好（递归），然后通过把数组中的数排序确定根节点的数值，再递归确定各子树中根节点的数值。这一题写递归写爽了。

AC代码：

#include<cstdio>
#include<stack>
#include<iostream>
#include<vector>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
struct Node{
 int value;
 int lhs;
 int rhs;
 int lhs_num;
 int rhs_num;
 Node(int v, int l, int r):value(v),lhs(l),rhs(r),lhs_num(0),rhs_num(0){}
 Node() :lhs_num(0), rhs_num(0){}
}buf[101];
int countChild(Node* root){//确定各节点的左右孩子数目
 if (root->lhs != -1){
  root->lhs_num = countChild(&buf[root->lhs]);
 }
 else root->lhs_num = 0;
 if (root->rhs != -1){
  root->rhs_num = countChild(&buf[root->rhs]);
 }
 return root->lhs_num + root->rhs_num + 1;
}
void build(Node* root, int num[]){//递归构建树，我写递归可是一个好手哈
 root->value = num[root->lhs_num];
 if (root->lhs_num > 0)
  build(&buf[root->lhs], num);
 if (root->rhs_num > 0)
  build(&buf[root->rhs], num + root->lhs_num + 1);
}
void seqTraversal(Node* root){//层序遍历，用作最后输出用
 queue<Node*>Q;
 Q.push(root);
 bool firstFlag = true;//用作格式化输出
 while (!Q.empty()){
  Node *front = Q.front();
  Q.pop();
  if (firstFlag){
   firstFlag = false;
   cout << front->value;
  }
  else{
   cout << " " << front->value;
  }
  if (front->lhs != -1)
   Q.push(&buf[front->lhs]);
  if (front->rhs != -1)
   Q.push(&buf[front->rhs]);
 }
 cout << endl;
}
int main(){
 freopen("F://Temp/input.txt", "r", stdin);
 int n;
 int num[100];
 cin >> n;
 for (int i = 0; i < n; ++i){
  cin >> buf[i].lhs >> buf[i].rhs;
 }
 for (int i = 0; i < n; ++i){
  cin >> num[i];
 }
 sort(num, num + n);
 countChild(buf);
 build(buf, num);
 seqTraversal(buf);
 return 0;
}

截图：



——Apie陈小旭