建树的时候,有时候没有必要大费周章地去通过结点构造一棵二叉树,我们利用各结点之间的数学关系,通过数组就可以实现一棵二叉树,假设结点序列为a,那么其左子就是a*2,右子就是a*2+1

由于二叉树中序遍历的结果是一串有序序列,那么我们可以通过中序来得到一棵二叉树

void leveltra(int root) {   //从根节点开始遍历
	if (root > n) {
		return;
	}
	leveltra(root * 2);  //左节点递归
	BST[root] = a[index++];
	leveltra(root * 2 + 1);  //右节点递归
}

//a表示有序序列,BST表示二叉搜索树

相关例题如下:

1064 Complete Binary Search Tree （30 分）

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.

A Complete Binary Tree (CBT) is a tree that is completely filled, with the possible exception of the bottom level, which is filled from left to right.

Now given a sequence of distinct non-negative integer keys, a unique BST can be constructed if it is required that the tree must also be a CBT. You are supposed to output the level order traversal sequence of this BST.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤1000). Then N distinct non-negative integer keys are given in the next line. All the numbers in a line are separated by a space and are no greater than 2000.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding complete binary search tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.

Sample Input:

10
1 2 3 4 5 6 7 8 9 0

Sample Output:

6 3 8 1 5 7 9 0 2 4

因为给出的是有序序列,所以我们可以以各个结点的位置序号为下标,来构造二叉树

如果按顺序输出结果,那么就是二叉树层序遍历的结果

实现代码如下：

#include <iostream>
#include <algorithm>
using namespace std;

int a[1010];
int levelTravese[1010];
int index = 1;   
int n = 0;

void leveltra(int root) {   //从根节点开始遍历
	if (root > n) {
		return;
	}
	leveltra(root * 2);  //左节点
	levelTravese[root] = a[index++];
	leveltra(root * 2 + 1);  //右节点
}

int main() {

	cin >> n;  //读入有多少个数
	for (int i = 1; i <= n; ++i) {
		cin >> a[i];
	}
	sort(a+1, a + n+1);   //进行排序
	leveltra(1);

	for (int j = 1; j <= n; ++j) {
		if (j == n) {
			cout << levelTravese[j];
		}
		else {
			cout << levelTravese[j] << " ";
		}
	}

	system("PAUSE");
	return 0;
}