1064 Complete Binary Search Tree (30 分)(dfs+层序遍历)

题目描述:

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
思路:这个完全二叉树和那个大根堆是一样的结构,所以知道了如果根节点从1开始编号,完全二叉树的左孩子就是他编号的二倍,有孩子是他编号的二倍+1,这样的话就能直接dfs建树
,而又因为BST树的排序性,所以将节点排序,在中序建树的时候按照从小到大给值就可以啦,层序遍历就不说啦

#include<iostream>
#include<string.h>
#include<cmath>
#include<set>
#include<map>
#include<string>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>
#include<algorithm>
#include<climits>
using namespace std;
typedef long long ll;
inline ll read() {
    ll sum = 0, f = 1;
    char p = getchar();
    for (; !isdigit(p); p = getchar()) if (p == '-')f = -1;
    for (; isdigit(p); p = getchar())  sum = sum * 10 + p - 48;
    return sum * f;
}
const int maxn = 10001;
int n;
int v[maxn],val[maxn];
int tree[maxn][2], tot;
int now=0;
void dfs(int u, int n) {//中序建树
    if (now * 2 <= n) {
        tree[u][0] = u * 2;
        dfs(tree[u][0], n);
    }
    val[u] = v[++now];//给值
    if (now * 2 + 1 <= n) {
        tree[u][1] = u * 2 + 1;
        dfs(tree[u][1], n);
    }
}
void levelorder(int u) {
    queue<int>Q;
    Q.push(u);
    bool f = 0;
    while (!Q.empty()) {
        int x = Q.front(); Q.pop();
        if (f)printf(" ");
        printf("%d", val[x]);
        f = 1;
        if (tree[x][0])Q.push(tree[x][0]);
        if (tree[x][1])Q.push(tree[x][1]);
    }
}
int main() {
    n = read();
    for (int i = 1; i <= n; i++) {
        v[i] = read();
    }
    sort(v + 1, v + n + 1);
    dfs(1, n);
    levelorder(1);
    return 0;
}