1099 Build A Binary Search Tree (30 分)

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.

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

分析

这一题与1064完全二叉树一题做法十分相似。
思路:
1.先构建一棵只有结点没有元素的树;
2.将数据输入进数组,数组从小到大排序;
3.找到根节点,并计算出左子树的结点个数,直接对应到数组,确定根节点元素;
4.更新左右范围和根节点值,对左子树和右子树进行分治。

代码

#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<

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