PAT 1099 Build A Binary Search Tree[BST性质]

1099 Build A 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.

Given the structure of a binary tree and a sequence of distinct integer keys, there is only one way to fill these keys into the tree so that the resulting tree satisfies the definition of a BST. You are supposed to output the level order traversal sequence of that tree. The sample is illustrated by Figure 1 and 2.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤100) which is the total number of nodes in the tree. The next N lines each contains the left and the right children of a node in the format left_index right_index, provided that the nodes are numbered from 0 to N−1, and 0 is always the root. If one child is missing, then −1 will represent the NULL child pointer. Finally N distinct integer keys are given in the last line.

Output Specification:

For each test case, print in one line the level order traversal sequence of that tree. All the numbers must be separated by a space, with no extra space at the end of the line.

Sample Input:

9
1 6
2 3
-1 -1
-1 4
5 -1
-1 -1
7 -1
-1 8
-1 -1
73 45 11 58 82 25 67 38 42

Sample Output:

58 25 82 11 38 67 45 73 42

题目大意：给定一棵二叉查找树的树结构，和一组不同的数，那么就能形成唯一的一棵BST，给出这棵BST的层次遍历顺序。

我的AC代码，一遍过，开心。

#include <iostream>
#include <algorithm>
#include <map>
#include<queue>
#include<vector>
using namespace std;
struct Node{
    int data,left,right;
}node[];

int d[];
vector<int> in;
queue<int> level;
void inorder(int root){
    if(node[root].left!=-)inorder(node[root].left);
    in.push_back(root);
    if(node[root].right!=-)inorder(node[root].right);
}
int main()
{
    int n;
    cin>>n;
    int left,right;
    for(int i=;i<n;i++){
        cin>>left>>right;
        node[i].left=left;
        node[i].right=right;
    }
    for(int i=;i<n;i++)
        cin>>d[i];
    sort(d,d+n);//从小到大排列即是中序遍历的结果。
    inorder();
    for(int i=;i<n;i++){
        node[in[i]].data=d[i];//安排进去了
    }
    //接下来层次遍历了。
    level.push();
    vector<int> lev;
    while(!level.empty()){
        int top=level.front();
        level.pop();
        lev.push_back(node[top].data);
        if(node[top].left!=-)level.push(node[top].left);
        if(node[top].right!=-)level.push(node[top].right);
    }
    cout<<lev[];
    for(int i=;i<n;i++){
        cout<<" "<<lev[i];
    }
    return ;
}

//猛一看觉得挺难，实际上要利用BST的性质就很好做了。 即中序遍历是数据从小到大排列的。

1.先对给出的二叉树结构，那么此时建树；

2.先对二叉树的下标进行中序遍历，并且将输入的数据从小到大排列，然后直接赋值就可以了。

3.再使用队列对其层次遍历即可。