1099. Build A Binary Search Tree (30)

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:

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

 1 #include<stdio.h>
 2 #include<math.h>
 3 #include<set>
 4 #include<algorithm>
 5 #include<vector>
 6 #include<queue>
 7 using namespace std;
 8
 9 struct node
10 {
11     int l,r,v;
12 };
13
14 node Tree[110];
15 vector<int> vv;
16 int cnt = 0;
17 void inOder(int root)
18 {
19     if(Tree[root].l != -1)
20         inOder(Tree[root].l);
21     Tree[root].v = vv[cnt++];
22     if(Tree[root].r != -1)
23         inOder(Tree[root].r);
24 }
25
26 int main()
27 {
28     int n,tem;
29     scanf("%d",&n);
30     for(int i = 0 ;i < n;++i)
31     {
32         scanf("%d%d",&Tree[i].l,&Tree[i].r);
33     }
34
35     for(int i = 0 ;i < n;++i)
36     {
37         scanf("%d",&tem);
38         vv.push_back(tem);
39     }
40     sort(vv.begin(),vv.end());
41     inOder(0);
42     queue<node> qq;
43     qq.push(Tree[0]);
44     bool fir = 1;
45     while(!qq.empty())
46     {
47         node ntem = qq.front();
48         qq.pop();
49         if(fir)
50         {
51             fir = 0;
52             printf("%d",ntem.v);
53         }
54         else
55         {
56             printf(" %d",ntem.v);
57         }
58         if(ntem.l != -1)
59             qq.push(Tree[ntem.l]);
60         if(ntem.r != -1)
61             qq.push(Tree[ntem.r]);
62     }
63     printf("\n");
64     return 0;
65 }