1099 Build A Binary Search Tree(30分)

タイトル

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.
figBST.jpg
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

ぶんせき

この問題は1064の完全な二叉樹の問題とよく似ている.考え方:1.まず、ノードだけで要素のない木を構築します.2.データを配列に入力し、配列を小さいものから大きいものに並べ替える.3.ルートノードを見つけ、左サブツリーのノード数を計算し、配列に直接対応し、ルートノード要素を決定する.4.左右範囲とルートノード値を更新し、左サブツリーと右サブツリーを分治する.

コード#コード#

#include
using namespace std;
struct BNode{
	int data;
	int left,right;
}BTree[120];
int GetSum(BNode BTree[],int root){
	if(root>=0){
		int n=0;
	if(BTree[root].left>=0){
		//cout<=0){
		//cout<=0&&left<=right){
	    int leftnum=GetSum(BTree,BTree[root].left);
	    BTree[root].data=vec[left+leftnum];
	    //cout<>n;
	for(int i=0;i>BTree[i].left>>BTree[i].right;
	}
	int a[n];
	for(int i=0;i>a[i];
	}
	sort(a,a+n);
	BuildBTree(BTree,a,0,n-1,0);
	queueq;
	q.push(0);
	int flag=0;
	while(!q.empty()){
		if(BTree[q.front()].left>0)//have a lchild
		{//cout<0)//have a rchild
		{//cout<