PAT甲级——1099 Build A Binary Search Tree (二叉搜索树)

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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

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

题目大意：将N个数放入一棵定型了的二叉树，使其满足二叉搜索树的性质。

思路：先将数据Data排好序，二叉树中存放数据的下标就行。

对于BST中的每个节点，它的key值对应的下标 index = 其上层节点传递过来的 M - 其右子树节点的个数 rightNum。若当前节点是其parent节点的左孩子，这个传递过来的M值就是parent节点的下标；若当前节点是parent节点的右孩子，那么M就是其parent节点的M。根节点的M值为N-1。

 1 #include <iostream>
 2 #include <vector>
 3 #include <queue>
 4 #include <algorithm>
 5 using namespace std;
 6 struct node {
 7     int left, right, 
 8         rightNum,
 9         index;
10 };
11 vector <node> tree;
12 vector <int> Data;
13 int getNum(int t);
14 void getIndex(int t, int M);
15 void levelOrder(int t);
16 int main()
17 {
18     int N;
19     scanf("%d", &N);
20     tree.resize(N);
21     for (int i = 0; i < N; i++)
22         scanf("%d%d", &tree[i].left, &tree[i].right);
23     Data.resize(N);
24     for (int i = 0; i < N; i++)
25         scanf("%d", &Data[i]);
26     sort(Data.begin(), Data.end());
27     getIndex(0, N - 1);
28     levelOrder(0);
29     return 0;
30 }
31 void levelOrder(int t) {
32     queue <int> Q;
33     Q.push(t);
34     while (!Q.empty()) {
35         t = Q.front();
36         Q.pop();
37         printf("%d", Data[tree[t].index]);
38         if (tree[t].left != -1)
39             Q.push(tree[t].left);
40         if (tree[t].right != -1)
41             Q.push(tree[t].right);
42         if (!Q.empty())
43             printf(" ");
44     }
45 }
46 void getIndex(int t, int M) {
47     if (t == -1) {
48         return;
49     }
50     tree[t].rightNum = getNum(tree[t].right);
51     tree[t].index = M - tree[t].rightNum;
52     getIndex(tree[t].left, tree[t].index - 1);
53     getIndex(tree[t].right, M);
54 }
55 int getNum(int t) {
56     if (t == -1)
57         return 0;
58     return getNum(tree[t].left) + getNum(tree[t].right) + 1;
59 }