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);
}