1099 Build A Binary Search Tree (30 point(s))

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.

figBST.jpg

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

做法同1064

#include<bits/stdc++.h>
using namespace std;
const int maxn=110;
int n,t=0;
int in[maxn];
struct  node{
    
    
    int lchild,rchild;
    int data;
}Node[maxn];
void inOrder(int root){
    
    
    if(root==-1) return;
    inOrder(Node[root].lchild);
    Node[root].data=in[t++];
    inOrder(Node[root].rchild);
}
void BFS(int root){
    
    
    queue<int> Q;
    Q.push(root);
    t=0;
    while(!Q.empty()){
    
    
        int now=Q.front();
        if(t!=0) printf(" ");
        t++;
        printf("%d",Node[now].data);
        Q.pop();
        if(Node[now].lchild!=-1) Q.push(Node[now].lchild);
        if(Node[now].rchild!=-1) Q.push(Node[now].rchild);
    }
}
int main(){
    
    
    scanf("%d",&n);
    for(int i=0;i<n;++i) scanf("%d %d",&Node[i].lchild,&Node[i].rchild);
    for(int i=0;i<n;++i) scanf("%d",&in[i]);
    sort(in,in+n);
    inOrder(0);
    BFS(0);
}