1099. Build A Binary Search Tree

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

按照输入建一个数，然后dfs依次赋值，最后bfs输出。

#include<iostream>#include<vector>#include<algorithm>#include<queue>using namespace std;typedef struct node{int num;int value;int l,r;};node a[100];int s;int va[100];queue <node> k;void dfs( int n){if(a[n].l!=-1)dfs(a[n].l);a[n].num=s;a[n].value=va[s];s++;if(a[n].r!=-1)dfs(a[n].r);return ;}int main(){s=0;int n;cin>>n;for(int i=0;i<n;i++){cin>>a[i].l>>a[i].r;}for(int i=0;i<n;i++)cin>>va[i];sort(va,va+n);dfs(0);k.push(a[0]);node p;bool flag =true;while(!k.empty()){p=k.front();k.pop();if(flag){flag=false;cout<<p.value;}elsecout<<" "<<p.value;if(p.l!=-1)k.push(a[p.l]);if(p.r!=-1)k.push(a[p.r]);}return 0;}