1064 Complete Binary Search Tree (30分)(BST和CST的性质)

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 (≤). 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


我们知道二叉搜索树有一个基本的性质，就是他的中序遍历结果就是所有元素从小到大排序的结果。
我们又知道该二叉树是完全二叉树，因此可根据中序遍历和完全二叉树这两个条件完全确定一棵BST。

直接将原数组排序然后从头到尾按照中序遍历建树。


这题需要总结的知识点：
　　BST中序遍历结果是所有元素从小到大排序的结果。
　　知道一棵树是完全二叉树且知道中序遍历结果可以唯一确定一棵树。
　　用数组建树时，数组顺序即时树的层序遍历结果。

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

const int maxn = 1000 + 5;
int val[maxn];

int tree[maxn << 2];

int cnt, n;

void build(int k) {
    if(k <= n) {
        build(k << 1);
        tree[k] = val[cnt ++];
        build(k << 1 | 1);
    }
}

int main() {
    scanf("%d", &n);
    for(int i = 0; i < n; i ++) {
        scanf("%d", &val[i]);
    }
    sort(val, val + n);
    build(1);
    for(int i = 1; i <= n; i ++) {
        if(i != 1) printf(" ");
        printf("%d", tree[i]);
    }
    printf("
");
    return 0;
}