1099 Build A Binary Search Tree (30 分)
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 Nlines 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<bits/stdc++.h>
using namespace std;
struct Node{
int data, lchild, rchild;
}node[110];
int n, wei[110], ind = 0;
void bct(int root){
if(node[root].lchild != -1)
bct(node[root].lchild);
node[root].data = wei[ind++];
if(node[root].rchild != -1)
bct(node[root].rchild);
}
void levelorder(int root){
queue<int> q;
q.push(root);
ind = 0;
while(!q.empty()){
int top = q.front();
q.pop();
printf("%d", node[top].data);
if(++ind < n)
printf(" ");
if(node[top].lchild != -1)
q.push(node[top].lchild);
if(node[top].rchild != -1)
q.push(node[top].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", &wei[i]);
sort(wei, wei + n);
bct(0);
levelorder(0);
return 0;
}