1099 Build A Binary Search Tree （30 分)

2019-04-17

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:

1	58 25 82 11 38 67 45 73 42

作者: CHEN, Yue

单位: 浙江大学

时间限制: 200 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出一颗二叉树的形状，再给出一组数填充进此树，要求输出这棵树的层序遍历。

分析

因为是二叉树，把输入的数排序之后即为中序遍历。中序遍历此树，过程中即可知道各个位置上对应的数。用一个map存储层号和各层上的数即可。

30分的题出的比20分的还简单，也是醉了。

代码

#include <iostream>
#include <vector>
#include <stack>
#include <algorithm>
#include <deque>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include <sstream>

using namespace std;
int n,cnt;
vector<int> tree[101];
vector<int> num;
map<int,vector<int>> layer;
void inorder(int root,int l){
    if(root==-1)
        return;
    inorder(tree[root][0],l+1);
    layer[l].push_back(num[cnt++]);
    inorder(tree[root][1],l+1);
}
int main() {
    cin>>n;
    num.resize(n);
    int l,r;
    for(int i=0;i<n;i++){
        cin>>l>>r;
        tree[i].push_back(l);
        tree[i].push_back(r);
    }
    for(int i=0;i<n;i++)
        cin>>num[i];
    sort(num.begin(),num.end());
    inorder(0,0);
    int f=0;
    for(auto it=layer.begin();it!=layer.end();it++){
        for(int x:it->second) {
            f++ == 0 ? cout << "" : cout << " ";
            cout<<x;
        }
    }
    return 0;
}