A1099. Build A Binary Search Tree

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

#include<cstdio>
#include<iostream>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
bool cmp(int a, int b){
    return a < b;
}
typedef struct NODE{
    int lchild, rchild;
    int key;
}node;
node tree[1001];
int N, num[1001], index = 0;
void inOrder(int root){
    if(root == -1)
        return;
    inOrder(tree[root].lchild);
    tree[root].key = num[index++];
    inOrder(tree[root].rchild);
}
void levelOrder(int root){
    int cnt = 0;
    queue<int> Q;
    if(root != -1){
        Q.push(root);
    }
    while(Q.empty() == false){
        int temp = Q.front();
        Q.pop();
        cnt++;
        if(cnt == N)
            printf("%d", tree[temp].key);
        else printf("%d ", tree[temp].key);
        if(tree[temp].lchild != -1)
            Q.push(tree[temp].lchild);
        if(tree[temp].rchild != -1)
            Q.push(tree[temp].rchild);
    }
}
int main(){
    scanf("%d", &N);
    for(int i = 0; i < N; i++){
        scanf("%d%d", &tree[i].lchild, &tree[i].rchild);
    }
    for(int i = 0; i < N; i++){
        scanf("%d", &num[i]);
    }
    sort(num, num + N, cmp);
    inOrder(0);
    levelOrder(0);
    cin >> N;
    return 0;
}

总结：

1、题意：给出一个二叉树的具体形状，给出一些键值，要求将这些键值按照给定的形状插入，使之成为搜索树。

2、二叉搜索树的中序序列是从小到大的有序序列。根据这一性质，先对序列进行排序，就得到了搜索树的中序序列。再对给出的二叉树进行中序遍历，在遍历的过程中插入keys，就得到了一个搜索树。