1064 Complete Binary Search Tree (PAT甲级)

这道题柳婼的解法很巧妙,理解上稍稍抽象,也就是当中序遍历时,level[r] = inorder[t++];

我按照她的思路写的代码如下:

#include 
#include 
#include 

int N, t;
std::vector inorder, level;

void in(int r){
    if(r >= N){
        return;
    }
    in(2 * r + 1);
    level[r] = inorder[t++];
    in(2 * r + 2);
}

int main(){
    scanf("%d", &N);
    inorder.resize(N);
    level.resize(N);
    for(int i = 0; i < N; ++i){
        scanf("%d", &inorder[i]);
    }
    sort(inorder.begin(), inorder.end());
    t = 0;
    in(0);
    for(int i = 0; i < N; ++i){
        printf("%d%s", level[i], i == N - 1 ? "\n" : " ");
    }
    return 0;
}

我一开始的解法如下,复杂很多,先是把loc按levelOrder排列出来,再用中序遍历把loc排列出来,再找loc与值之间一一对应的关系,最后再把值levelOrder列出来。

#include 
#include 
#include 
#include 

int N, pivot;
std::vector vec, loc, inorderVec;
std::map left, right, mp;

void inorder(int r){
    if(left[r] != 0){
        inorder(left[r]);
    }
    inorderVec.push_back(r);
    if(right[r] != 0){
        inorder(right[r]);
    }
}

int main(){
    scanf("%d", &N);
    vec.resize(N);
    for(int i = 0; i < N; ++i){
        scanf("%d", &vec[i]);
    }
    sort(vec.begin(), vec.end());
    loc.push_back(1);
    pivot = 0;
    while(loc.size() != N){
        loc.push_back(loc[pivot] << 1);
        left[loc[pivot]] = loc[pivot] << 1;
        if(loc.size() != N){
            loc.push_back(loc[pivot] << 1 | 1);
            right[loc[pivot]] = loc[pivot] << 1 | 1;
        }
        ++pivot;
    }
    inorder(1);
    for(int i = 0; i < N; ++i){
        mp[inorderVec[i]] = vec[i];
    }
    for(int i = 0; i < N; ++i){
        printf("%d%s", mp[loc[i]], i == N - 1 ? "\n" : " ");
    }
    return 0;
}

题目如下:

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

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