1099. Build A Binary Search Tree (30)

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本文介绍了一种构建二叉搜索树的方法，通过给定的二叉树结构和一组不同的整数键，唯一地填充这些键以满足二叉搜索树的定义，并实现了层序遍历输出。

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1099. Build A Binary Search Tree (30)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

分析：

（1）给出“固定结构”的二叉搜索树，然后给出树中的数组，要求还原这棵树；

（2）将数组从小到大排列正好是二叉搜索树的中序遍历结果，因此可以借此来还原整棵树；

（3）树还原之后，就可以利用队列轻松的进行层序遍历了；

这道题目关键还是在于上述的第二点。

写个中序遍历即可，只不过这次不是输出而是往里面填数，很不错的一道题

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> val;
int cnt = 0;
struct node{
    int val;
    int left;
    int right;
};
vector<node> tree;
void inorder(int root){
    if(root != -1){
        inorder(tree[root].left);
        tree[root].val = val[cnt++];
        inorder(tree[root].right);
    }
}
void levelorder(int root){
    queue<node> q;
    vector<node> ans;
    q.push(tree[root]);
    while(!q.empty()){
        node tmp = q.front();
        q.pop();
        ans.push_back(tmp);
        if(tmp.left != -1) q.push(tree[tmp.left]);
        if(tmp.right != -1) q.push(tree[tmp.right]);
    }
    for(int i = 0; i < ans.size(); i++){
        if(i == 0) printf("%d",ans[i].val);
        else printf(" %d",ans[i].val);
    }
    puts("");
}
int main(){
    int n;
    cin >> n;
    tree.resize(n);
    for(int i = 0; i < n; i++){
        cin >> tree[i].left >> tree[i].right;
    }
    val.resize(n);
    for(int i = 0; i < n; i++){
        cin >> val[i];
    }
    sort(val.begin(),val.end());
    inorder(0);
    levelorder(0);
    return 0;
}