PAT 1064 Complete Binary Search Tree (30 分)

• 題目：

  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.

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.

• 題目大意

  給出一個BST樹，并且符合完全二叉樹的定義，按照層次周遊輸出這顆二叉樹的節點的值

• 解題思路

  按照完全二叉樹的定義，用數組存儲節點的值（從1開始）,則2x為x節點的左孩子，2x+1為x節點的右孩子。由于BST樹，中序周遊有序，隻用中序周遊完全二叉樹輸入節點的值即可（輸入的值存儲在數組中，需要按照從小到大排序）。最後隻用從前向後輸出數組的值，就是這顆樹層次周遊的值

• 代碼實作：

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
const int MAX = 10010;
int node[MAX];
int N, k = 0;
vector<int> num;
void mid_order(int root){    //中序周遊，輸入節點的值
    if(root > N){
        return;
    }
    mid_order(2 * root);
    node[root] = num[k];
    k++;
    mid_order(2 * root + 1);
}
int main()
{
    scanf("%d", &N);
    for(int i = 0; i < N; i++){
        int temp;
        scanf("%d", &temp);
        num.push_back(temp);
    }
    sort(num.begin(), num.end());                      //排序
    mid_order(1);
    for(int i = 1; i <= N; i++){                           //從前往後輸出完全二叉樹的值
        if(i != 1){
            printf(" %d", node[i]);
        } else
            printf("%d", node[1]);
    }
    return 0;
}