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:

91 62 3-1 -1-1 45 -1-1 -17 -1-1 8-1 -173 45 11 58 82 25 67 38 42

Sample Output:

58 25 82 11 38 67 45 73 42

思路：该题与上一题1064差不多，题目给出每个根结点对应左右子树（数组下标），因此在模拟中序遍历的时候直接代入即可

之后按层序遍历输出，用队列的方式输出就行了

#include<stdio.h>#include<stdlib.h> #include<math.h>#include<string.h>#include<algorithm>#include <vector> #include<stack> #include <queue>  #include<map>#include<iostream>  #include <functional> #define MAX 110#define MAXD 10001#define TELNUM 10using namespace std;struct node{int data;int l;int r;}Node[MAX];int n, k = 0;int num[MAX];int num1[MAX];void tian(int root){if(root > n - 1)return;if(Node[root].l != -1){tian(Node[root].l);}Node[root].data = num[k++];if(Node[root].r != -1){tian(Node[root].r);}}void BFS(int root){queue<int> q;int t = 0;q.push(root);while(!q.empty()){int temp = q.front();q.pop();printf("%d",Node[temp].data);t++;if(t < n){printf(" ");}if(Node[temp].l != -1)q.push(Node[temp].l);if(Node[temp].r != -1)q.push(Node[temp].r);}}int main(void){scanf("%d",&n);for(int i = 0; i < n; i++){scanf("%d%d",&Node[i].l,&Node[i].r);}for(int i = 0; i < n; i++){scanf("%d",&num[i]);}sort(num,num + n);tian(0);BFS(0);return 0;}